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Full text of "University algebra :"

UC-NRLF 






EH MEMOEIAM 
Edward Bright 





(Brcenlcafs Jttathematical 3ttk^ 



UNIVERSITY ALGEBRA. 



DESIGNED FOR THE USE OF SCHOOLS 
AND COLLEGES. 



PREPARED BY 



WEBSTER WELLS, S. B., 

i 

ASSISTANT PROFESSOR OF MATHEMATICS IN THE MASSACHUSETTS 
INSTITUTE OF TECHNOLOGY; 




LEACH, SHEWELL, AND SANBORN, 
BOSTON AND NEW YORK. 






COPYRIGHT, 1880. 



q/v- 



By WEBSTER WELL?.. 



PREFACE. 



This work was designed to take the place of Green- 
leaf's Higher Algebra, portions of which have been used 
in the preparation of the present volume. It contains 
the topics usually taught in High Schools and Colleges, 
and the author's aim has been to present the subject in a 
compact form and in clear and concise language. The 
principles have been developed with regard to logical ac- 
curacy, and care has been given to the selection of exam- 
ples and practical illustrations which should exercise the 
student in all the common applications of the algebraic 
analysis. The full treatment given in the earlier chap- 
ters renders the previous study of a more elementary 
text-book unnecessary. 

Attention is invited to the following chapters, including 
those in which the most important changes have been 
made in the Higher Algebra : — 

Parentheses. 

Factoring. 

Zero and Infinity. 

Theory of Exponents. 

Simultaneous Equations involving Quadratics. 

Binomial Theorem for Positive Integral Exponents. 

Undetermined Coefficients. 

Logarithms. 

The answers have been put by themselves in the back 
part of the book, and those have been omitted which, if 



7979-5 



i / PREFACE. 

given, would destroy the utility of the problem. The ex- 
amples are over eighteen hundred in number, and are pro- 
gressive, commencing with simple applications of the 
rules, and passing gradually to those which require some 
thought for their solution. 

The works of Todhunter and Hamblin Smith, and other 
standard volumes, have been consulted in the preparation 
of the work, and have furnished a number of examples 
and problems. The author has also received numerous 
suggestions from practical teachers, to whom he would 
here express his thanks. 

WEBSTER WELLS. 

Boston, 1884. 



UNIVERSITY ALGEBRA. 



CONTENTS. 



Chapter Paob 

I. Definitions and Notation 1 

Symbols of Quantity 1 

Symbols of Operation 2 

Symbols of Relation 4 

Symbols of Abbreviation 5 

Algebraic Expressions 5 

Axioms 8 

Negative Quantities 12 

II. Addition 14 

III. Subtraction 19 

IV. Use of Parentheses 21 

V. Multiplication 24 

VI. Division 31 

VII. Formula 38 

VIII. Factoring 40 

IX. Greatest Common Divisor 53 

X. Least Common Multiple 61 

XI. Fractions 66 

Reduction of Fractions 70 

Addition and Subtraction of Fractions 78 

Multiplication of Fractions 82 

Division of Fractions 85 

Complex Fractions 87 

XII. Simple Equations. — One unknown quantity 89 

Transformation of Equations 91 

Solution of Equations 95 



Vlll 



CONTENTS. 



XIII. Problems. — One unknown quantity 103 

XIV. Simple Equations. — Two unknown quantities 113 

Elimination 115 

XV. Simple Equations. — More than two unknown quan- 
tities 123 

XVI. Problems. —More than one unknown quantity 127 

Generalization of Problems 133 

XVII. Discussion of Problems 137 

Interpretation of Negative Results 139 

XVIII. Zero and Infinity 142 

Problem of the Couriers 143 

XIX. Inequalities 148 

XX. Involution 153 

Involution of Monomials 153 

Involution of Polynomials 154 

Square of a Polynomial 155 

Cube of a Binomial 156 

Cube of a Polynomial 157 

XXI. Evolution 158 

Evolution of Monomials 159 

Square Root of Polynomials 160 

Square Root of Numbers 163 

Cube Root of Polynomials 168 

Cube Root of Numbers 171 

Any Root of Polynomials 174 

XXII. The Theory of Exponents 176 

XXIII. Radicals 188 

Reduction of Radicals 1 88 

Addition and Subtraction of Radicals 193 

Multiplication of Radicals 194 

Division of Radicals 196 

Involution of Radicals 197 

Evolution of Radicals 198 

Reduction of Fractions with Irrational Denominators 199 

Imaginary Quantities 202 

Quadratic Surds 205 

Radical Equations 208 



CONTENTS. ix 

XXIV. Quadratic Equations. — One unknown quantity... 210 

Pure Quadratic Equations 211 

Affected Quadratic Equations 213 

XXV. Problems. — Quadratic Equations. — One unknown 

quantity 223 

XXVI. Equations in the Quadratic Form 227 

XXVII. Simultaneous Equations involving Quadratics 233 

XXVIII. Problems. — Quadratic Equations. — Two unknown 

QUANTITIES 244 

XXIX. Theory of Quadratic Equations 249 

Discussion of the General Equation 249 

XXX. Discussion of Problems leading to Quadratic 

Equations 257 

Interpretation of I maginary Results 259 

Problem of the Lights 259 

XXXI. Ratio and Proportion 262 

XXXII. Variation 270 

XXXIII. Arithmetical Progression 274 

XXXIV. Geometrical Progression 282 

XXXV. Harmonical Progression 291 

XXXVI. Permutations and Combinations 294 

XXXVII. Binomial Theorem. — Positive Integral Exponent 298 

XXXVIII. Undetermined Coefficients 304 

Expansion of Fractions into Series 307 

Expansion of Radicals into Series 310 

Decomposition of Rational Fractions 312 

Reversion of Series 318 

XXXIX. Binomial Theorem. —Any Exponent 321 

XL. Summation of Infinite Series 328 

Recurring Series 328 

The Differential Method 332 

Interpolation 336 



x CONTENTS. 

XLI. Logarithms 339 

Properties of Logarithms 342 

Use of the Table 348 

Solutions of Arithmetical Problems by Logarithms 354 

Exponential Equations 358 

Application of Logarithms to Problems in Compound In- 
terest 359 

Exponential and Logarithmic Series 362 

Arithmetical Complement 366 

XLII. General Theory of Equations 369 

Divisibility of Equations 370 

Number of Roots .- 371 

Formation of Equations 373 

Composition of Coefficients 374 

Fractional Roots 376 

Imaginary Roots 376 

Transformation of Equations 377 

Descartes' Rule of Signs 383 

Derived Polynomials 385 

Equal Roots 386 

Limits of the Roots of an Equation 389 

Sturm's Theorem 392 

XLIII. Solution of Higher Numerical Equations 399 

Commensurable Roots 400 

Recurring or Reciprocal Equations 404 

Cardan's Method for the Solution of Cubic Equations. .... 408 

Biquadratic Equations 411 

Incommensurable Roots 412 

Horner's Method 412 

Approximation by Double Position 417 

Newton's Method of Approximation 419 

Answers to Examples 421 

Table of the Logarithms of Numbers from 1 to 10,000. .Appendix. 



ALGEBRA. 



I. — DEFINITIONS AND NOTATION. 

1. Quantity is anything that can be measured ; as dis- 
tance, time, weight, and number. 

2. The Measurement of quantity is accomplished by find- 
ing the number of times it contains another quantity of the 
same kind, assumed as a standard. This standard is called 
the unit of measure. 

3. Mathematics is the science of quantities and their re- 
lations. 

4. Algebra is that branch of mathematics in which the 
relations of quantities are investigated, and the reasoning 
abridged and generalized, by means of symbols. 

5. The Symbols employed in Algebra are of four kinds: 
symbols of quantity, symbols of operation, symbols of relation, 
and symbols of abbreviation. 

SYMBOLS OF QUANTITY. 

6. The Symbols of Quantity generally used are the 
figures of Arithmetic and the letters of the alphabet. 

The figures are used to represent known quantities and 
determined values, and the letters any quantities whatever, 
known or unknown. 

7. Known Quantities, or those whose values are given, 



■2 ALGEBRA. 

when not expressed by figures, are usually represented by the 
first letters of the alphabet, as a, b, c. 

8. Unknown Quantities, or those whose values are not 
given, are usually represented by the last letters of the 
alphabet, as x, y, z. 

9. Zero, or the absence of quantity, is represented by the 
symbol 0. 

10. Quantities occupying similar relations in different op- 
erations are often represented by the same letter, distinguished 
by different accents, as a', a", a'", read "a prime," "a second," 
"a third," etc. ; or by different subscript figures, as a 1} a 2) a s , 
read " a one/' " a two," " a three," etc. 

SYMBOLS OF OPERATION. 

11. The Symbols of Operation are certain signs or char- 
acters used to indicate algebraic operations. 

12. The Sign of Addition, + , is called "plus." Thus, 
a -(- b, read " a plus b" indicates that the quantity b is to be 
added to the quantity a. 

13. The Sign of Subtraction, — , is called "minus." 
Thus, a — b, read " a minus b" indicates that the quantity 
b is to be subtracted from the quantity a. 

The sign ~ indicates the difference of two quantities when 
it is not known which of them is the greater. Thus, a ~ b 
indicates the difference of the two quantities a and b. 

14. The Sign of Multiplication, x , is read " times" 
"into," or "multiplied by." Thus, aXb indicates that the 
quantity a is multiplied by the quantity b. 

A simple point (. ) is sometimes used in place of the sign X- 
The sign of multiplication is, however, usually omitted, except 
between two arithmetical figures separated by no other sign; 
multiplication is therefore indicated by the absence of any 
sign. Thus, 2ab indicates the same as 2 X a X b, or 2 . a . b. 



DEFINITIONS AND NOTATION. 3 

15. The quantities multiplied are called factors, and the 
result of the multiplication is called the product. 

16. The Sign of Division, ~, is read "divided by." 
Thus, a -4- b indicates that the quantity a is divided by the 
quantity b. 

Division is otherwise often indicated by writing the divi- 
dend above, and the divisor below, a horizontal line. Thus, 

- indicates the same as a -i- b. Also, the sign of division 

b 

may be replaced in an operation by a straight or curved line. 

Thus, a I b, or b ) a, indicates the same as a -f- b. 

17. The Exponential Sign is a figure or letter written at 
the right of and above a quantity, to indicate the number of 
times the quantity is taken as a factor. Thus, in x 3 , the 3 in- 
dicates that x is taken three times as a factor ; that is, x 3 is 
equivalent to xxx. 

The product obtained by taking a factor two or more times 
is called a power. A single letter is also often called the 
first power of that letter. Thus, 

a 2 is read "a to the second power," or "a square," and 
indicates a a; 

a 3 is read " a to the third power," or " a cube," and indi- 
cates acta; 

a 4 is read " a to the fourth power," or " a fourth," and indi- 
cates a a a a; 

a n is read " a to the nth power," or " a nth," and indicates 
a a a etc., to n factors. 

The figures or letters used to indicate powers are called 
exponents ; and when no exponent is written, the first power 
is understood. Thus, a is equivalent to a 1 . 

The root of a quantity is one of its equal factors. Thus, 
the root of a 2 , a 3 , or a* is a. 

18. The Radical Sign, \f , when prefixed to a quantity, 
indicates that some root of the quantity is to be extracted. 



4 ALGEBRA. 

Thus, 

SJ a indicates the second or square root of a; 

^]a indicates the third or cuhe root of a; 

$ a indicates the fourth root of a; and so on. 

The index of the root is the figure or letter written over 
the radical sign. Thus, 2 is the index of the square root, 3 of 
the cube root ; and so on. 

When the radical sign has no index written over it, the 
index 2 is understood. Thus, sj a is the same as tf a. 

SYMBOLS OF RELATION. 

19. The Symbols of Relation are signs used to indicate 
the relative magnitudes of quantities. 

20. The Sign of Equality, =, read " equals" or "equal 
to," indicates that the quantities between which it is placed 
are equal. Thus, x = y indicates that the quantity x is equal 
to the quantity y. 

A statement that two quantities are equal is called an 
equation. Thus, x -\- 4=2 x — 1 is an equation, and is read 
" x plus 4 equals 2x minus 1." 

21. The Sign of Ratio, : , read " to," indicates that the 
two quantities between which it is placed are taken as the 
terms of a ratio. Thus, a : b indicates the ratio of the quan- 
tity a to the quantity b, and is read " the ratio of a to b." 

A proportion, or an equality of ratios, is expressed by writ- 
ing the sign =, or the sign : :, between equal ratios. Thus, 
30 : 6 = 25 : 5 indicates that the ratio of 30 to G is equal to 
the ratio of 25 to 5, and is read "30 is to 6 as 25 is to 5." 

22. The Sign of Inequality, > or < , read " is greater 
than" or "is less than" respectively, when placed between 

two quantities, indicates that the quantity toward which the 
opening of the sign turns is the greater. Tims, x > y is 
read "x is greater than y" ; x—6< y is read "x minus 6 
is less than y." 



DEFINITIONS AND NOTATION. 5 

23. The Sign of Variation, cc, read "varies as," indicates 
that the two quantities between which it is placed increase 

or diminish together, in the same ratio. Thus, a oc _ is read 
" a varies as c divided by d." 



SYMBOLS OF ABBREVIATION. 

24. The Signs of Deduction, .-. and v , stand the one for 
therefore or hence, the other for since or because. 

25. The Signs of Aggregation, the vinculum , the 

bar | , the parenthesis ( ) , the brackets [ ] , and the braces j £, 

indicate that the quantities connected or enclosed by them are 
to be subjected to the same operations. Thus, 

a + b X oc, a x, (a + b) x, [a + fr]x, \a + bc x, 



b 



all indicate that the quantity a + b is to be multiplied 
by x. 

26. The Sign of Continuation, , stands for and so 

on, or continued by the some law. Thus, 

a, a + b, a + 2 b, a + 3 b, is read 

"a, a plus b, a plus 2b, a plus 3 b, and so on." 

ALGEBRAIC EXPRESSIONS. 

27. An Algebraic Expression is any combination of alge- 
braic symbols. 

28. A Coefficient of a quantity is a figure or letter pre- 
fixed to it, to show how many times the quantity is to be 
taken. Thus, in 4«, 4 is the coefficient of a, and indicates 
that a is taken four times, or a + a + a + a. Where any 
number of quantities are multiplied together, the product of 



6 ALGEBEA. 

any of them may be regarded as the coefficient of the product 
of the others; thus, in abed, ab is the coefficient of ed, 
b of ac d, a b d of c, and so on. 

When no coefficient of a quantity is written, 1 is understood 
to he the coefficient. Thus, a is the same as 1 a, and x y is 
the same as 1 x y. 

29. The Terms of an algebraic expression are its parts 
connected by the signs + or — . Thus, 

a and b are the terms of the expression a + b ; 

2 a, b 2 , and — 2 a c, of the expression 2 a + b 2 — 2 a c. 

30. The Degree .of a term is the number of literal factors 
which it contains. Thus, 

2 a is of the first degree, as it contains but one literal factor, 
a & is of the second degree, as it contains two literal factors. 

3 a b 2 is of the third degree, as it contains three literal factors. 

The degree of any term is determined by adding the expon- 
ents of its several letters. Thus, a b 2 c s is of the sixth degree. 

31. Positive Terms are those preceded by a plus sign ; as, 

+ 2 a, or + a b 2 . 

When a term has no sign written, it is understood to be 
positive. Thus, a is the same as + a. 

Negative Terms are those preceded by a minus sign ; as, 

— 3 a, or —be. 

This sign can never be omitted. 

32. In a positive term, the coefficient indicates how many 
times the quantity is taken additively (Art. 28) ; in a nega- 
tive term, the coefficient indicates how many times the quan- 
tity is taken svbtractively. Thus, 

+ 2 x is the same as + x + x ; 

— 3 a is the same as — a — a — a. 



DEFINITIONS AND NOTATION. 7 

33. If the same quantity be both added to and subtracted 

from another, the value of the latter will not be changed ; 

hence if any quantity b be added to any other quantity a, and 

b be subtracted from the result, the remainder will be a; 

that is, 

# a + b — b = a. 

Consequently, equal terms affected by unlike signs, in an 
expression, neutralize each other, or cancel. 

34. Similar or Like Terms are those which differ only in 
their numerical coefficients. Thus, 

2 x y 2 and — 1 xif are similar terms. 

Dissimilar or Unlike Terms are those which are not similar. 
Thus, 

b x 2 y and bxy 2 are dissimilar terms. 

35. A Monomial is an algebraic expression consisting of 
only one term ; as, 5 a, 7 a b, or 3 b' 2 c. 

A monomial is sometimes called a simple quantity. 

36. A Polynomial is an algebraic expression consisting of 
more than one term ; as, a + b, or 3 a 2 + b — 5 b 3 . 

A polynomial is sometimes called a compound quantity, or a 
multinomial. 

37. A Binomial is a polynomial of two terms ; as, 

a — b, 2 a + b 2 , ov dad 2 — b. 

A binomial whose second term is negative, as a — b, is some- 
times called a residual. 

38. A Trinomial is a polynomial of three terms ; as, 

a + b + c, or a b + c 2 — b 3 . 

39. Homogeneous Terms are those of the same degree ; as, 

a 2 , 3 be, and — 4 x 2 . 



$ ALGEBRA. 

40. A polynomial is homogeneous when all its terms are 
homogeneous ; as, a 3 + 2 a b c — 3 b 3 . 

41. A polynomial is said to be arranged according to the 
decreasing powers of any letter, when the term having the 
highest exponent of that letter is placed first, that having 
the next lower immediately after, and so on. Thus, 

a 3 +3a 2 b + 3ab 2 + b 3 

is arranged according t6 the decreasing powers of a. 

A polynomial is said to be arranged according to the increas- 
ing powers of any letter, when the term having the lowest 
exponent of that letter is placed first, that having the next 
higher immediately after, and so on. Thus, 

a 3 + 3 a 2 b + 3 a b 2 + b 3 

is arranged according to the increasing powers of b. 

42. The Reciprocal of a quantity is 1 divided by that 
quantity. Thus, the reciprocal of 

a is - , and of x + y is 



a x + y 

43. The Interpretation of an algebraic expression consists 
in rendering it into an arithmetical quantity, by means of the 
numerical values assigned to its letters. The result is called 
the numerical value of the expression. 

Thus, the numerical value of 

4d+ 3 be — d 
when a = 4, b = 3, c = 5, and d = 2, is 

4x4 + 3x3x5-2 = 16 + 45 - 2 = 59. 

AXIOMS. 

44. An Axiom is a self-evident truth. 

Algebraic operations are based upon definitions, and the 
following axioms : — 



DEFINITIONS AND NOTATION. 9 

1. If the same quantity, or equal quantities, be added to 
equal quantities, the sums will he equal. 

• 2. If the same quantity, or equal quantities, he subtracted 
from equal quantities, the remainders will he equal. 

3. If equal quantities be multiplied by the same quantity, or 
by equal quantities, the products will be equal. 

4. If equal quantities be divided by the same quantity, or 
by equal quantities, the quotients will be equal. 

5. If the same quantity be both added to and subtracted from 
another, the value of the latter will not be changed. 

6. If a quantity be both multiplied and divided by another, 
the value of the former will not be changed. 

7. Quantities which are equal to the same quantity are equal 
to each other. 

8. Like powers and like roots of equal quantities are equal. 

9. The whole of a quantity is equal to the sum of all its 
parts. 

EXERCISES ON THE PRECEDING DEFINITIONS AND 

PRINCIPLES. 

45. Translate the following algebraic expressions into 
ordinary language : 



d 



m 



1. 3 a 2 + b c — q. 5. cd : — = ab : \J x*. 



3 



n 



x 



2. 4 m . 6. (a — b)x = [c + d~] y. 

<- -} 3a — d 

" 2c + b 



3. ^ a + b = ^ a 2 — c. 7. {m + r — s}n = 



4,mn>pa. 8. \J -^- < (e - d) (h + fj. 

46. Put into the form of algebraic expressions the follow- 
ing : 

1. Five times a, added to two times b. 

2. Two times x, minus y to the second power. 



10 ALGEBRA. 

3. The difference of x and y. 

4. The product of «, b, c square, and d cube. 

5. x + y multiplied by a — b. 

6. a square divided by the sum of b and c. 

7. x divided by 3, increased by 2, equals three times y, 
diminished by 11. 

8. The reciprocal of a + b, plus the square of a } minus the 
cube root of b, is equal to the square root of c. 

9. The ratio of 5 a divided by b, to d divided by c square, 
equals the ratio of x square y cube to y square z fourth. 

10. The product of in and a + b is less than the reciprocal 
of x cube. 

11. The product of x + y and x — y is greater than the 
product of the square of a — d into the cube of a + b. 

12. The quotient of a divided by 3 a — 2 is equal to the 
square root of the quotient of m + n divided by 2x — y 2 . 

47. Find the numerical values of the following : — 
When a = 6, b = 5, c = 4, and d — 1, of 

1. a 2 + 2 a b — c + d. 4. a 2 (a + b) — 2abc. 

2. 2a?-2a 2 b + c 3 . 5. 5a 2 b-±ab 2 + 21c. 

3. 2a 2 + 3bc-5. 6. 7 a 2 + (a- b) (a-c). 

When a = 4, b = 2, c = 3, and d = l, of 



a 



2 J,2 



1.15a-7(b+c)-d. 10. ^ + | + ^. 

8. 25a 2 -7(b 2 + c 2 ) + d\ 11. 4 + 1 . 

„« £» c '25 a — 30 c— d 

9. - H 1- - . 12. . 

bed b + c 

When a = \, b = \, c = \, and x = 2, of 

IS. (2 a + 3 b + 5 c) ($ a + 3 b - 5 e) (2 a- 3 b + 15 c). 






DEFINITIONS AND NOTATION. H 

15. x* - (2 a + 3 b) x 3 + (3 a - 2 b) x 2 - ex + be. 
When a=b, and b = £, of 

16 5 a + ^ - C 3 a - ( 2 a - ^)] 

17 13 a + 3 b + {7 Q + b) + [3 a + 8 (4 a - b)~\) 

2a + Sb 

When b = 3, c = 4, d = 6, and e = 2, of 

18. V 27T- v' 27+ y/2Z 19. V 3^7+ ^"9^-^27. 

When « = 16, & = 10 ; cc = 5, and y=l, of 

20. (b - x) (y/a + b) + ^ (a - b) (x + y). 

48. What is the coefficient of 

1. x in 3 n 2 x ? 3. x y in — 20 m? xyz z ? 

2. a c s in a J 2 c 3 d 4 ? 4. ra 2 w 3 in 5 a 8 m 2 a; ?i 3 ? 

What is the degree of 

5. 3ax? 6. 2m*nx*? 7. a 2 b s c 2 d 5 ? 8. 2mcr 2 y 3 .?? 

Arrange the following expressions according to the increas- 
ing powers of x : 

9. 2cc 2 -3a; + x 3 + l-4a; 4 . 

10. 3 x y* — 5 x s y + y i — x 4 — x 2 y 2 . 

Arrange the following expressions according ' o the decreas- 
ing powers of a : 

11. 1- a 2 -2 a + a 3 + 2 a*. 

12. aJ 3 -i 4 + a 4 -4« 2 i 2 -3a 3 5. 



12 ALGEBRA. 



NEGATIVE QUANTITIES. 

49. The signs + and — , besides indicating the operations 
of addition and subtraction, are also used, in Algebra, to indi- 
cate the nature or quality of the quantities to which they are 
prefixed. 

To illustrate, let us suppose a person, having a property of 
$ 500, to lose $ 150, then gain $ 250, and finally to incur a debt 
of $ 450 ; it is required to find the amount of his property. 

Since gains have an additive effect on property, and debts or 
losses a subtractive effect, we may indicate these different 
qualities algebraically by prefixing the signs + and — to them, 
respectively ; thus, we should represent the transactions as 
follows, 

$ 500 - $ 150 + $ 250 - $ 450 ; 

which reduces to $ 150, the amount required. 

But suppose, having a property of $ 500, he incurs a debt 
of $ 700 ; we should represent the transaction algebraically as 

follows, 

$500-1700; 

or, as incurring a debt of $ 700 is equivalent to incurring two 
debts, one of $ 500 and the other of $ 200, the transaction may 
be expressed thus, 

$ 500 - $ 500 - $ 200. 

Now since, by Art. 33, $ 500 and — $ 500 neutralize each 
other, we have remaining the isolated negative quantity 
— $200 as the algebraic representative of the required prop- 
erty. In Arithmetic, we should say that he owed or was in 
debt $200; in Algebra, we make also the equivalent state- 
ment that his property amounts to —$200. 

In this way we can conceive the possibility of the indepen- 
dent existence of negative quantities; and as, in Arithmetic, 
losses may be added, subtracted, multiplied, etc., precisely as 
though they were gains, so, in Algebra, negative quantities 



DEFINITIONS AND NOTATION. 13 

may be added, subtracted, multiplied, etc., precisely as though 
they were positive. 

The distinction of positive and negative quantities is applied 
in a great many cases in the language of every-day life and in 
the mathematical sciences. Thus, in the thermometer, we 
speak of a temperature above zero as +, and one below as — ; 
for instance, +25° means 25° above zero, and —10° means 
10° below zero. In navigation, north latitude is considered 
-j-, and south latitude — ; longitude west of Greenwich is con- 
sidered +, and longitude east of Greenwich — ; for example, 
a place in latitude — 30°, longitude + 95°, would be in latitude 
30° south of the equator, and in longitude 95° west of Green- 
wich. And, in general, when we have to consider quantities 
the exact reverse of each other in quality or condition, we 
may regard quantities of either quality or condition as posi- 
tive, and those of the opposite quality or condition as negative. 
It is immaterial which quality we regard as positive ; but hav- 
ing assumed at the commencement of an investigation a certain 
quality as positive, we must retain the same notation through- 
out. 

The absolute value of a quantity is the number represented 
by that quantity, taken independently of the sign affecting it. 
Thus, 2 and — 2 have the same absolute value. 

But as we consider a person who owns $ 2 as better off 
than one who owes $2, so, in Algebra, we consider + 2 as 
greater than — 2 ; and, in general, any positive quantity, 
however small, is considered greater titan any negative quan- 
tity. 

Also, as we consider a person who owes $ 2 as better off than 
one who owes $3, so, in Algebra, we consider — 2 as greater 
than —3; and, in general, oftivo negative quantities, that is 
regarded as the greater which has the less number of units, or 
which has the smaller absolute value. 

Again, as we consider a person who has no property or debt 
as better off than one who is in debt, so, in Algebra, zero is 
considered greater than any negative quantity. 



14 ALGEBRA. 



II. — ADDITION. 

50. Addition, in Algebra, is the process of collecting two 
or more quantities into one equivalent expression, called, the 
sum. 

51. In Arithmetic, when a person incurs a debt of a certain 
amount, we regard his property as diminished by the amount 
of the debt. So, in Algebra, using the interpretation of nega- 
tive quantities as given in Art. 49, adding a negative quantity 
is equivalent to subtracting an equal positive quantity. Thus, 
the sum of a and — b is obtained by subtracting b from a, giv- 
ing as a result a — b. 

Hence, the addition of monomials is indicated by uniting 
the quantities with their respective signs. Thus, the sum of 
a, — b, c, d, — e, and — f, is 

a—b+c+d—e —f. 

The addition of polynomials is indicated by enclosing them 
in parentheses (Art. 25), and uniting the results with + signs. 
Thus, the sum of a + b and c — d is 

(a + b) + (c — d). 

52. Let it be required to add c — d to a + b. 

If we add c to a + b, the sum will be a + b + c. But we 
have to add to a + b a quantity which is d less than c. Conse- 
quently our result is d too large. Hence the required sum will 
be a + b + c diminished by d, or a + b + c — d. 

Hence, the addition of polynomials may also be indicated by 
uniting their terms with their respective signs. 

53. Let it be required to add 2 a and 3 a. 
By Art. 32, 2 a = a + a, 

and 3 a = a + a + a. 



ADDITION. 15 

Hence (Art. 52) the sum of 2 a and 3 a is indicated by 

a -\- a + a + a + a, 
which, by Art. 32, is equal to 5 a. Hence, 2a + 3a = 5a. 

54. Let it be required to add — 3 a and — 2 a. 
By Art. 32, — 3 a = — a — a — a, 

and — 2 a = — a — a. 

Hence (Art. 52), the sum of — 3 a and — 2 a is indicated by 

— a — a — a — a — a, 

or — 5 a (Art. 32). Hence, — 3 a — 2 a = — 5 a. 

From our ideas of negative quantities (Art. 49), we may ex- 
plain this result arithmetically as follows : 

If a person has two debts, one of $ 3 and the other of $ 2, 
he may be considered to be in debt to the amount of $ 5. 

55. Let it be required to add 4 a and — 2 a. 

4 a = a + a+ a + a, 

and — 2« = — a — a. 

Hence, the sum of 4 a and — 2 a is indicated by 

a + a + a + a — a — a. 

Now, by Art. 33, the third and fourth terms are neutralized 
by the fifth and sixth, leaving as the result a + a, or 2 a. 
Hence, 4 « — 2 a = 2 a. 

We may explain this result arithmetically as follows : 
If a person has $4 in money, and incurs a debt of $2, his 
property may be considered to amount to $ 2. 

56. Let it be required to add — 4 a and 2 a. 

— 4:a = — a — a — a — a, 
and 2a = ffl+«. 



16 ALGEBRA. 

Hence, the sum of — 4 a and 2 a is indicated by 

— a — a — a — a + a + a. 

The third and fourth terms neutralize the fifth and sixth, 
leaving as the result — a — a or —2a. Hence, 

— 4ta + 2a = — 2a. 

We may explain this result arithmetically as follows : 
If a person has $ 2 in money, and incurs a debt of 84, he 
may be considered to be in debt to the amount of $2. 

.57. From Arts. 55 and 56 we derive the following rule 
for the addition of two similar (Art. 34) terms of opposite 
sign: 

To add two similar terms, the one positive and the other 
negative, subtract the smaller coefficient from the larger, affix 
to the result the common symbols, and prefix the sign of the 
larger. 

For example, the sum of 7 x y and — 3xy is 4:xy, 

the sum of 3 a 2 b s and — 11 a~ b 3 is — 8 a 2 b 3 . 

58. In Arithmetic, when adding several quantities, it 
makes no difference in which order we add them ; thus, 
3 + 5 + 9, 5 + 3 + 9, 9+3 + 5, etc., all give the same result, 
17. So also in Algebra, it is immaterial in what order the 
terms are united, provided each has its proper sign. Thus, 

— b + a is the same as a — b. 

Hence, in adding together any number of similar terms, 
some positive and some negative, we may add the positive 
terms first, and then the negative, and finally combine these 
two results by the rule of Art. 57. 

Thus, in finding the sum of 2 a, — a, la, 6 a, — 4 a, and 

— 5 a, the sum of the positive terms 2 a, 7 a, and 6 a, is 15 a, 
and the sum of the negative terms — a, — 4 a, and — 5 a, 
is — 10 a ; and the sum of 15 a and — 10 a is 5 a. 

59. Let it be required to add 6 a — 7 x, 3 x — 2 a + 3 y, 
and 2 x — a — mn. 



ADDITION. 1 7 

We might obtain the sum in accordance with Art. 52, by- 
uniting the terms by their respective signs, and combining 
similar terms by the methods previously given. It is however 
customary in practice, and more convenient, to set the expres- 
sions down one underneath the other, similar terms being in 
the same vertical column ; thus, 

6 a — 1 x 
— 2 a + 3 x + 3 y 
— a + 2 x —mn 



3 a — 2 x + 3 y — m n. 

It should be remembered that only similar terms can be 
combined by addition ; and that the algebraic sum of dissimilar 
terms can only be indicated by uniting them by their respective 
signs. 

60. From the preceding principles and illustrations is de- 
rived the following 

RULE. 

To add together two or more expressions, set them down one 
underneath the other, similar terms being in the same vertical 
column. Find the sum of the similar terms, and to the result 
obtained unite the dissimilar terms, if any, by their respective 
signs. 







EXAMPLES. 






1. 


2. 


3. 


4. 


5. 


la 


— 6 m 


13 n 


— 4 a x 


2a 2 b 


3 a 


m 


n 


— 3 ax 


-a-b 


a 


— 11 m 


— 20n 


a x 


11 a- b 


5 a 


— 5m 


6 n 


— lax 


-5a 2 b 


11a 


— m 


8n 


— ax 


±a 2 b 


a 


20 m 


— n 


12 ax 


-9aH 



18 ALGEBRA. 

6. 7. 8. 

la — mp 2 2 a — 3 x ab + c d 

a + 6 mp 2 — a + 4x — ab -\- cd 

— 11 a — 3 mp 2 a + x 3 a b — 2 cd 

8 a + 11 m j9 2 5 a — 7 a; lab — 5 cd 

— 9 a— 1 mp 2 —4: a— x —4tab + 6cd 

18 a — 15 mp 2 —3 a + 1 x 2 ab — 5cd 



Find the sum of the following : 

9. 4,xy z, — 3xy z, — 5xy z, 6x y z, — 9xy z, and 3 x y z. 

10. 5 m n 2 — 8x 2 y, — m n 2 + x 2 y, — 6m n 2 — 3x 2 y, 4:mn 2 
+ 1 x 2 y, 2 m n 2 + 3 x 2 y, and — ra ri 2 — 2x 2 y. 

11. 3a 2 + 2ab + 4,b 2 , 5a 2 -Sab + b 2 , -a 2 +5ab-b 2 , 
18 a 2 -20 ab- 19 b 2 , and 14 a 2 - 3 a b + 20 b 2 . 

12. 2a — 5b — c +1, 3b — 2-6a + 8c, c + 3a-4, and 
1 + 2 b - 5 c. 

13. 6x — 3y+lm, 2 n — x + y, 2 y — 4x— 5 m, and 
m + n — y. 

14. 2 a - 3 b + 4 <7, 2 & - 3 d + 4 c, 2 cZ - 3 e + 4 a, and 
2c-3a + 4i. 

15. 3 cc — 2 y — z, 3 y — 5 x — 1 z, 8 z — y — x, and 4 x. 

16. 2 m — 3n + 5r — t, 2 n — 6 t — 3 r — m, r + 3 m — 5n 
+ 4t, and 3 t — 2 r + 1 n — 4 m. 

17. 4:inn + 3 ab — 4 c, 3 x — 4 a b + 2 m n, and 3 m 2 — 4 p. 

18. 3 a + b — 10, c — d — a, —4c + 2a — 3b — l, and 
4 a; 2 + 5 - 18 m. 

19. 4a;8_5 a 8_ 5ax 2 +6a 2 x, 6a s + 3x* + Aax 2 + 2a 2 x, 
-11 X s +19 ax 2 - 15 a 2 x, and 10 x 3 + 1 a 2 x + 5 a 3 - 18 a x 2 . 

20. la — 5if, S^x + 2a, oif — \/x, and — 9a + 7tfx. 

21. 3 a b + 3 (« + b), - a b + 2 (a + b), 7 a b — 4 (a + b), 
and — 2 a b + 6 (a + b). 

22. lsjy-4(a-b), 6 \J y + 2 (a - b), 2 ^ y + (a-b), and 
sjy — 3(a — b). 



SUBTRACTION. 19 



III. — SUBTRACTION. 

61. Subtraction, in Algebra, is the process of finding one 
of two quantities, when their sum and the other quantity are 
given. 

Hence, Subtraction is the converse of Addition. 

The Minuend is the sum of the quantities. 

The Subtrahend is the given quantity. 

The Remainder is the required quantity. 

As the remainder is the difference between the minuend 
and subtrahend, subtraction may also be defined as the process 
of finding the difference between two quantities. 

62. Subtraction may be indicated by writing the subtra- 
hend after the minuend, with a — sign between them. Thus, 
the subtraction of b from a is indicated by 

a — b. 

In indicating subtraction in this way, the subtrahend, if a 
negative quantity or a polynomial, should be enclosed in a 
parenthesis. Thus, the subtraction of —b from a is indi- 
cated by 

and the subtraction of b — c from a by 

a— (b — c). 

63. Let it be required to subtract b — c from a. 
According to the definition of Art. 61, we are to find a 

quantity which when added to b — c will produce a ; this 
quantity is evidently a — b + c, which is the remainder re- 
quired. 

Now, if we had changed the sign of each term of the sub- 
trahend, giving — b + c, and had added the resulting expres- 
sion to a, we should have arrived at the same result, a — b + c. 



20 ALGEBRA. 

Hence, to subtract one quantity from another, we may change 
the sign of each term of the subtrahend, and add the residt to 
the minuend. 

64. 1. Let it be required to subtract 3 a from 8 a. 
According to Art. 63, the result may be obtained by adding 

— 3 a to 8 a, giving 5 a (Art. 55). 

2. Subtract 8 a from 3 a. 

By Art. 63, the result is 3 a — S a or —5a (Art. 56). 

3. Subtract — 2 a from 3 a. 
Result, 3 a + 2 a or 5 a. 

4. Subtract 3 a from —2 a. 

Result, — 2a-3«or —5 a. 

5. Subtract — 2 a from —5 a. 
Result, — 5 a + 2 a or —3 a. 

6. Subtract —5a from —2 a. 
Result, — 2 a + 5 a or 3 a. 

65. In Arithmetic, addition always implies augmentation, 
and subtraction diminution. In Algebra this is not always 
the case ; for example, in adding — 2 a to 5 a the sum is 3 a, 
which is smaller than 5 a ; also, in subtracting —2 a from 5 a 
the remainder is 7 a, which is larger than 5 a. Thus, the 
terms Addition, Subtraction, Sum, and Remainder have a 
much more general signification in Algebra than in Arith- 
metic. 

66. From Art. 63 we derive the following 

RULE. 

To subtract one expression from another, set the subtrahend 
underneath the minuend, similar terms being in the same ver- 
tical column. Change the sign of each term of the subtrahend 
from + to — , or from — to + , and add the restdt to the 
minuend. 



USE OF PAKENTHESES. 21 







EXAMPLES. 






1. 


2. 


3. 


4. 


5. 


27 a 


17 x 


-13 3, 


- 10 m n 


5 a 2 b 


13 a 


-11b 


Ay - 


- 18 m n 


UaH 



ab + cd — ax 7x+5y—3a 

Aab — 3 cd-\- Aax x—7y+ha—A 

8. 9. 

7 abc-llx + 5y- 48 5 \Ja- 3 y 2 + 7 a -6 

llabc+ 3x + 7?/ + 100 3<Ja+ y 2 -5a-7 

10. Subtract —5b from - 12 J. 

11. From 31 x 2 - 3 y 2 + a b take 17 x 2 + 5 if - 4 a b + 7. 

12. Subtract a — b + c from a + b — c. 

13. Subtract 6a — 3^ — 5c from Qa-\-3b — 5 c + 1. 

14. From 3m-5ft+r-2s take 2 r + 3 n — m — 5 s. 

15. Take 4 a — b + 2 i — 5 d from a" — 3 & + a - c. 

16. From m 2 + 3 n z take — 4 m' 2 — 6 ?i 8 + 71 cc. 

17. From a + b take 2«-25 and — a + b. 

18. From a — b — c take — a + b + c and a — b + c. 



IV. — USE OF PARENTHESES. 

67. The use of parentheses is very frequent in Algebra, 
and it is necessary to have rules for their removal or introduc- 
tion 



22 ALGEBRA. 

68. Let it be required to indicate the addition of 3 a. and 
5 b — c + 2 d ; this we may do by placing the latter expression 
in a parenthesis, prefixing a + sign, and writing after the 
former quantity, thus : 

3 a + (5 b - c + 2 d). 

If the operation be performed, we obtain (Art. 60), 

3a + 5b — c + 2d. 

69. Again, let it be required to indicate the subtraction of 
5b — c-\-2d from 3 a ; this we may do by placing the former 
expression in a parenthesis, prefixing a — sign, and writing 
after the latter quantity, thus : 

3a-(5b-c + 2d). 

If the operation be performed, we obtain (Art. 66), 

3 a — 5 b + c — 2d. 

70. It will be observed that in the former case the signs 
»f the terms within the parenthesis are unchanged when the 
parenthesis is removed ; while in the latter case the sign of 
each term within is changed, from + to — , or from — to +. 
Hence, we have the following rule for the removal of a paren- 
thesis : 

If the parenthesis is preceded by a -f- sign, it may be re- 
moved if the sign of every enclosed term be unchanged; and 
if the parenthes-is is preceded by a — sign, it may be removed 
if the sign of every enclosed term be changed. 

71. To enclose any number of terms in a parenthesis, we 
take the reverse of the preceding rule : 

Any number of terms may be enclosed in a parenthesis, with 
a + sign prefixed, if the sign of every term enclosed be un- 
changed ; and in a parenthesis, with a — sig?i prefixed, if 
the sign of every term enclosed be changed. 



USE OF PARENTHESES. 23 

72. As the bracket, brace, and vinculum (Art. 25) have the 
same signification as the parenthesis, the rules for their re- 
moval or introduction are the same. It should be observed 
in the case of the vinculum, that the sign apparently prefixed 
to the first term underneath is in reality the sign of the vin- 
culum ; thus, + a — b signifies + (a — V), and — a — b signi- 
fies — (a — b). 

73. Parentheses will often be found enclosing others ; in 
this case they may be removed successively, by the preceding 
rule ; and it is better to begin by removing the inside pair. 

74. 1. Remove the parentheses from 3 a — (2 a — 5) — 

(-a + 7). 

Result, 3a — 2a + 5 + a — 7 — 2a — 2. 

2. Remove the parentheses etc., from 



6 a - [3 a + (2 a - { 5 a - [4 a - a - 2] } )]. 

In accordance with Art. 73, we remove the vinculum first, 
and the others in succession. Thus, 



6 a — [3 a +(2 a -{5 a -[4 a- a- 2]})] 
= 6 a - [3 a + (2 a - {5 a - [4 a - a + 2]})] 
= 6 a — [3 a + (2 a— {5 a — 4 a + a — 2})] 

= 6 a — [3 a + (2 a — 5 a + 4 a — a + 2)] 

= Qa — [3a + 2a — 5a + 4a — a + 2^\ 

= 6a — 3a — 2a + 5a — 4:a + a — 2 = 3a — 2, Ans. 

3. Enclose the last three terms of a — b~ c + d+ e —f in 
a parenthesis with a — sign prefixed. 

Result, a — b — c—(—d — e+f). 



24 ALGEBRA. 

EXAMPLES. 

Remove the parentheses, etc., from the following : 

4. a — (b — c) + (d — e). 

5. 3a-(2a-{a + 2}). 

6. 5 x — (2 x — 3 y) — (2 x +. 4 y). 

7. a — b + c — {a + b — c) — {c + b — a). 

8. in' — 2n+ (a — n + 3 ?m 2 ) — (5 a + 3 w — m 2 ). 

9. 2 m - [n — {3 wi -(2w- m) } ]. 



10. 8x — (5x — [4 a; — ?/ — &■]) — (— a; — 3 y). 



11. 2«-[5J+ {3c-(a+[2ft-3a + 4c])}]. 

12. 3c+(2a-[5c-{3a + c-4a}-]). 



13. 6 a - [5 a - (4 a - { - 3 a - [2 a- a- 1]})]. 



14. 2 m - [3 m - (5 m - 2) - { m - (2 wi -3m + 4)}]. 

75. As another application of the rule of Art. 70, we have 
the following four results : 

+ (+ a) is equivalent to + a ; 
+ ( — a) is equivalent to — a ; 

— (+ a) is equivalent to — a ; 

— (—a) is equivalent to + a. 



V. — MULTIPLICATION. 

76. Multiplication, in Algebra, is the process of taking 
one quantity as many times as there are units in another 
quantity. 

The Multiplicand is the quantity to be multiplied or taken. 

The Multiplier is the quantity by which we multiply. 

The Product is the result of the operation. 

The multiplicand and multiplier are often called factors. 



MULTIPLICATION. 25 

77. The product of the factors is the same, in whatever 
order they are taken. 

For we know, from Arithmetic, that the product of two 
numbers is the same, in whatever order they are taken ; thus 
we have 3 X 4 or 4 X 3 eacli equal to 12. Similarly, in Alge- 
bra, where the symbols represent numbers, we have a X b or 
b X a each equal to a b (Art. 14). 

78. Let it be required to multiply a — b by c. 

By Art. 77, multiplying a — b by c is the same as multiply- 
ing c by a — b. To multiply c by a — &, we multiply it first 
by a, and then by b, and subtract the second result from the 
first, e multiplied by a gives a c, and multiplied by b gives 
b c. Subtracting the second result from the first we have 

a c — b c 
the product required. 

79. Let it be required to multiply a — b by c — d. 

To multiply a — b by c — d, we multiply it first by c, and 
then by d, and subtract the second result from the first. By 
Art. 78, a — b multiplied by c gives ac — bc, and multiplied 
by d gives ad — bd. Subtracting the second result from the 
first, we have 

ac — bc — ad+bd 

the product required. 

80. We observe in the result of Art. 79, 

1. The product of the positive term a by the positive term 
c gives the positive term a c. 

2. The product of the negative term —b by the positive 
term c gives the negative term —be. 

3. The product of the positive terra a by the negative term 
— d gives the negative term — a d. 

4. The product of the negative tern> —b by the negative 
term — d gives the positive term b d. 



26 ALGEBRA. 

From these considerations we can state what is known as 
the Rule of Signs in Multiplication, as follows: 

+ multiplied by +, and — multiplied by — , produce + ; 
+ multiplied by — , and — multiplied by + , produce — . 

Or, as may he enunciated for the sake of hrevity with regard 
to the product of any two terms, 

Like signs produce + , and unlike signs produce — . 

81. Let it he required to multiply 7 a by 2 b. 

Since (Art. 77) the factors may he written in any order, we 
have 7ax2b = 7x2xaXb = 14:ab. Hence, 

The coefficient of the product is equal to the product of the 
coefficients of the factors. 

82. Let it he required to multiply a 3 hy a 2 . 

By Art. 17, a s means sX«X« } and a 2 means aXa; hence, 

a ! Xfl 2 = «X»X«X«X(i=« 5 . Hence, 

The exponent of a letter in the product is equal to the sum 
of its exponents in the factors. ' 

Or, in general, a m X a n = a m + n . 

83. In Multiplication we may distinguish three cases. 

CASE I. • 

84. WJien both factors are monomials. 

From Arts. 80, 81, and 82 is derived the following rule for 
the product of any two monomials. 

RULE. 

Multiply the numerical coefficients together ; annex to the 
residt the letters of both monomials, giving to curb letter an 
exponent equal to the sum of its exponents in the factors. Make 
the product + when the two factors have the same sign, and — 
when they have different signs. 



MULTIPLICATION. 27 

EXAMPLES. 

1. Multiply 2 «" by 3 a 2 . 

2« 4 x3a 2 = 6a 6 , Ans. 

2. Multiply a 3 b 2 c by - 5 a 2 b d. 

a 8 b 2 c X — 5 a 2 b d — — 5 a 5 b s c d, Ans. 

3. Multiply — 7 x m by — 5 se n . 

— 7 a; m x — 5 x n = 35 a: m+n , ^4?zs. 

4. Multiply 3 a (a; — y) 2 by 4« 3 (x- ?/). 

3 a (a - y) 2 X 4 a 3 (x-y) = 12 a 4 (a; - y) 3 , Ans. 

Multiply the following : 

5. 15 m 5 w 6 by 3 m n. 12. — 12 a 2 x by — 2 a 2 y. 

6. 3 a 6 by 2 a e. 13. 3 a m 6 n by — 5 a n b r . 

7. 17 a b c by — 8 a b c. 14. — 4 x m ?/" by — x n y n z b . 
8.-17 a 4 c 2 by - 3 a 2 c 2 . 15. 2 a m 5" by 5 a 3 b. 

9. 11 n 2 y by — 5 w 6 «. 16. — 7 m n x 2 by m n # r y 2 . 

10. 4a 6 by3aiy 2 . 17. 2 m (a - b) 2 by m (a - b). 

11. — 6 a b 2 c by a 3 b m. 18. 7 a (x — y) hy —3 a 2 b (x — y). 

19. Find the continued product of 8 a x 2 , 2 a 3 y, and 4 a; 3 v/ 4 . 

20. Find the continued product of 2 a c 2 , — 4 a c 3 , and 

-3«J 2 . 

CASE II. 

85. Wlien one of the factors is a polynomial. 
From Art. 78 we have the following 

RULE. 

Multiply each term of the multiplicand by the multiplier, 

remembering that like signs produce +, and unlike signs pro- 
duce — . 



28 ALGEBRA. 

EXAMPLES. 

1. Multiply 3 x — y by 2 x y. 

3 x — y 
2 x y 



6 x' 2 y — 2 x y 2 , Ans. 
2. Multiply 3 a — 5 x by — 4 ra. 

3 a — 5 x 
— 4 m 



,3 



.2 



— 12 a m + 20 ra #, ^4?is. 
Multiply tbe following : 

3. x 2 -2z-3by 4z. 7. -x 4 -10a; 3 + 5by-2x- 

4. 8 a 2 6 c - fZ by 5 a f/ 2 . 8. a 2 + 13 a& - 6 6 2 by 4 a b 2 

5. 3 x 2 + 6 x — 7 by — 2 x s . 9. ra 2 + m n + ?r by m n. 

6. 3 ra 2 — 5 ra ?i — ?i 2 by — 2 m. 10. 5 — 6 a — 8 a 3 by — 6 <x 

11. 5a; 3 -4x 2 -3z-2by-6r\ 

12. « 3 - 3 a 2 b + 3 a b 2 - b* by a 2 b 2 . 



CASE III. 

86. When both of the factors are polynomials. 

In Art. 79 we sbowed that tbe product of a — b and c — d 
rnigbt be obtained by multiplying a — b by c, and then by d, 
and subtracting tbe second result from tbe first. It would 
evidently be equally correct to multiply a — b by c, and then 
by — d, and add the second result to the first. On this we 
base the following rule for finding the product of two poly- 
nomials. 

RULE. 

Multiply each term of the multiplicand by each term of the 

multiplier, remembering that like signs 'produce +, and unlike 
signs produce —, and add the partial, products. 



MULTIPLICATION. 29 



EXAMPLES. 

1. Multiply 3 a - 2 b by 2 a- 5b. 

3a -2b 
2a —5b 



6 a 2 — Aab 

- 15 a b + 10 b 2 

6 a 2 — 19 a b + 10 6' 2 , ^ws. 

The reason for shifting the second partial product one place 
to the right, is that in general it enables us to place like terms 
in the same vertical column, where they are more conveniently 
added. 

2. Multiply x 2 + 1 — x 3 — x by x + 1. 

1 — X + X 2 — X 3 

1 + x 



1 — X + X 2 — X 3 

+ X — X 2 + X 3 — X* 

1 —x 4 , Ans. 

It is convenient, though not essential, to have both multi- 
plicand and multiplier arranged in the same order of powers 
(Art. 41), and to write the product in the same order. 

Multiply the following : 

3. 3 x 2 — 2 x y — y 2 by 2 x — Ay. 

4. x 2 + 2x + lhyx 2 -2x + 3. 

5. a + b — c by a — b + c. 

6. 3a-2bhy-2a + 4.b. 

7. a 2 + b 2 + ab by b — a. 

8. 1 + x + x 3 + x 2 by a x — a. 

9. 5 a 2 - 3 a b + 4 b 2 by 6 a - 5 b. 
10. 3 x 2 — 7 x + 4 by 2 x 2 + 9 x — 5. 



30 ALGEBRA. 

11. 6 x - 2 x- - 5 - a- 3 by x 2 + 10 - 2 x. 

12. 2a 3 +5a> 2 -8a:-7by4-5 : * ; -3a: 2 . 

13. a 3 b - a 2 b 2 - 4 a 6 8 by 2 a 2 5 - a b 2 . 

14. x m + 2 ij — 3xy n ~ l by 4 *"• + 5 y 2 — 4 a 4 y n . 

15. 6 a: 4 - 3 x 3 - a: 2 +6 a; -2 by 2x 2 + x + 2. 

16. m 4 — m 3 w + rn 2 n 2 — m w 3 + % 4 by m + n. 

17. ft 3 -3« 2 H3ai 2 - 6 3 by a 2 -2ab + b\ 

87. It is sometimes sufficient to indicate the product of 
polynomials, by enclosing each of the given factors in a paren- 
thesis, arid writing them one after the other, with or without 
the sign X between the parentheses. When the indicated 
multiplication is performed, the expression is said to be ex- 
panded or developed. 

1. Indicate the product of 2 x 2 — 3 x y + 6 by 3 x 2 + 3 x y — o. 
Kesult, (2x 2 -3xy + 6) (3x 2 +3xy-5). 

EXAMPLES. 

2. Expand (3 a + 4 b) (2 a + b). 

3. Expand (a* — a 3 x + a 2 x 2 — a x 3 + x 4 ) (a + x). 

4. Develop (a* — a: 4 ) X (« 4 — a,' 4 ). 

5. Develop (a m — a") (2 a — a n ). 

6. Expand (1 + x) (1 + a- 4 ) (1 — a- + .r 2 - a- 3 ). 

7. Find the value of (« + 2 a) (« — 3 x) (a + 4 x). 

8. Expand [« (a 2 — 3 a + 3) - 1] x [a (a - 2) + 1]. 

88. From the definition of Art. 76, X a means taken 
a times. Since taken any number of times produces 0, it 
follows that x a = 0. That is, 

If zero be multiplied by any quantity} the product is equal 
to zero. 



DIVISION. 31 

89. Since (+ a) X (+ b) —ab, and (— a) X (— &) =ab, 
it follows that in the indicated product of two factors, all the 
signs of both factors may be changed without altering the value 
of the expression. Thus, 

(x — y) {a — b) is equal to {y — x) (J> — a). 

Similarly we may show that in the indicated product of any 
number of factors, any even number of factors may In/re their 
signs changed without altering the value of the expression. 

Thus, (x — y) (c — d) (e —f) (g — h) is equal to 
{y-x) (c-d) (f-e) (g- h), or to 
(y — x) (d — c) {f-e) (h — g), etc.; but is not equal to 
(y-x){d~c)(f-e)(g-h). 



VI. — DIVISION. 

90. Division, in Algebra, is the process of finding one of 
two factors, when their product and the other factor are given. 

Hence, Division is the converse of Multiplication. 

The Dividend is the product of the two factors. 
The Divisor is the given factor. 
The Quotient is the required factor. 

91. Since the quotient multiplied by the divisor produces 
the dividend, it follows, from Art. 80, that if the divisor and 
quotient have the same sign, the dividend is + ; and if they 
have different signs, the dividend is — . Hence, 

+ divided by +, and — divided by — , produce + ; 
+ divided by — , and — divided by +, produce — . 
Hence, in division as in multiplication, 

Like signs produce +, and unlike signs produce — . 



32 ALGEBRA. 

92. Let it be required to find the quotient of 14 a b divided 
by 7 a. 

Since the quotient is such a quantity as when multiplied by 
the divisor produces the dividend, the quotient required must 
be such a quantity as when multiplied by 7 a will produce 
14 a b. That quantity is evidently 2 b. Hence, 

The coefficient of the quotient is equal to the coefficient of 
the dividend divided by the coefficient of the divisor. 

93. Let it be required to find the quotient of a 5 divided 
by a 3 . 

The quotient required must be such a quantity as when 
multiplied by a 3 will produce a 5 . That quantity is evidently 
a 2 . Hence, 

The exponent of a letter in the quotient is equal to its expo- 
nent in the dividend diminished by its exponent in the divisor. 

Or, in general, a m -f- a n = a m ~ n . 

94. If we apply the rule of Art. 93 to finding the quotient 
of a™ divided by a m , we have a m -f- a m = a m ~ m = a . 

Now, according to the previously given definition of an ex- 
ponent (Art. 17), a° has no meaning, and we are therefore at 
liberty to give to it any definition we please. As a m — a m = 1, 
we should naturally define a° as being equal to 1 ; and as a 
may represent any quantity whatever, 

Any quantity whose exponent is is' equal to 1. 

By this notation, the trace of a letter which has disappeared 
in the operation of division may be preserved. Thus, the 
quotient of a 2 b 3 divided by a 2 b 2 , if important to indicate that 
a originally entered into the term, may be written a b. 

95. In Division we may distinguish three cases. 

CASE I. 

96. When both dividend and divisor are monomials. 
From the preceding articles is derived the following 



DIVISION. 33 

RULE. 

Divide the coefficient of the dividend by that of the divisor ; 
and to the result annex the letters of the dividend, each with an 
exponent equal to its exponent in the dividend diminished by 
its exponent in the divisor; omitting all letters whose expo- 
nents become zero. Make the quotient + when the dividend and 
divisor have the same sign, and — when they have different 
signs. 

EXAMPLES. 

1. Divide 9 a 2 b c x y by 3 a b c. 

9a 2 bcxy-i-3abc = 3axy, Ans. 

2. Divide 24 a 4 m 3 n 2 by — 8 a m 3 n. 

24 a 4 m 3 n 2 -. — 8 a m 3 n — — 3 a 3 n, Ans. 

3. Divide — 35 x m by — 7 x n . 

— 35 x m -^- — 7x n = 5 x m ~ n , Ans. 

Divide tbe following : 

4. 12 a 5 by 4 a. 8. - 65 a 3 b 3 c 3 by - 5 a b 2 c 3 . 

5. 6 a 2 c by 6 a c. 9. 72 m° n by — 12 m 2 . 

6. 14 m 3 n 4 by - 7 m n 3 . 10. - 144 c 5 <P e 6 by 36 c 2 d 3 e. 

7. -18x 2 y 5 zhy9x 2 z. 11. - 91 x 4 y 3 z 2 by - 13 x 3 y 1 . 

CASE II. 

97. When the dividend is a polynomial and the divisor is 
a monomial. 

Tbe operation being just tbe reverse of that of Art. 85, we 
have the following 

RULE. 

Divide each term of the dividend by the divisor, remembering 
that like signs produce +, and tinlike signs produce — . 



34 ALGEBRA. 

EXAMPLES. 

1. Divide 9 a 3 ft + 6 a 4 e — 12 a ft by 3 a. 

3«)9a 3 H6 a 4 c — 12 a 5 
3a 2 i + 2« 3 c-4J, ^4»s. 

Divide the following : 

2. 8 a 3 6 c + 16 a 5 6 c — 4 a 2 c 2 by 4 a 2 c. 

3. 9 a 5 ft c - 3 a 2 ft + 18 a 3 ft c by 3 a ft. 

4. 20 a 4 ft c + 15 a 6 d 3 - 10 a 2 ft by - 5 a ft. 

5. 3 a 3 (a - ft) + 9 a (a + ft) by 3 a. 

6. 15 (x + y) 2 — 5 a (x + y) + 10 ft (x + y) by — 5 (x + y). 

7. 4 x 7 - 8 a- 6 - 14 t> + 2 a 4 - 6 x 3 by 2 x\ 

8. 9 a 4 + 27 x 3 - 21 a 2 by -3 a; 2 . 

9. _ a 6 £6 C 4 _ a 4 &5 c 3 + 3 a 8 £4 ^2 fcy _ ft 8 p ^ 

10. — 12 aP ft? - 30 a 12 ft 3 + 108 a" ft n by - 6 a m ft'". 

CASE III. 

98. When the divisor is a polynomial. 

1. Let it be required to divide 12 + 10 x 3 — 11 x — 21 x 2 by 
2z 2 -4-3x. 

We are then to find a quantity which when multiplied by 
2 x 2 - 4 - 3 x will produce 12 + 10 x 3 - 11 x - 21 x 2 . 

Now, in the product of two polynomials, the term containing 
the highest power of any letter in the multiplicand, multiplied 
by the term containing the highest power of the same letter 
in the multiplier, produces the term containing the highest 
power of that letter in the product. Hence, if the term con- 
taining the highest power of x in the dividend, 10 x 3 , be di- 
vided by the term containing the highest power of x in the 
divisor, 2 x 2 , the result, 5 x, will he the term containing the 
highest power of x in the quotient. 



DIVISION. 35 

Multiplying the divisor by 5 x, the term of the quotient 
already found, and subtracting the result, 10 x 3 — 20 x — 15 x 2 , 
from the dividend,' the remainder, 12 + 9 x — 6 x' 2 , may be re- 
garded as the product of the divisor by the rest of the quotient. 

Therefore, to find the rest of the quotient, we proceed as be- 
fore, regarding 12 + 9 x — 6 x 2 as a new dividend, and divid- 
ing the term containing the highest power of x, — 6 x' 2 , by 
the term containing the highest power of x in the divisor, 2 x 2 , 
giving as a result — 3, which is the term containing the high- 
est power of x in the rest of the quotient. 

Multiplying the divisor by — 3, the term of the quotient 
just found, and subtracting the result, — 6 x 2 + 12 + 9 x, from 
the second dividend, there is no remainder. Hence, 5 x — 3 
is the quotient required. 

99. It will be observed that in getting the terms of the 
quotient, we search for the terms containing the highest power 
of some letter in the dividend and divisor. These may be 
obtained most conveniently by arranging both dividend and 
divisor in order of powers commencing with the highest 
(Art. 41) ; this, too, facilitates the subsequent subtraction. 
We also should arrange each remainder or new dividend in 
the same order. 

It is customary to arrange the work as follows : 



10 x 3 - 21 x 2 - 11 x + 12 
10 x s - 15 x 2 - 20 x 



2 x 2 — 3 x — 4, Divisor. 
5 x — 3, Quotient. 



— 6x 2 + 9 a: + 12 

- 6x 2 + 9 a; + 12 



100. We might have obtained the quotient by dividing the 
term containing the lowest power of x in the dividend, 12, by 
the term containing the lowest power of x in the divisor, — 4, 
which would have given as a result — 3, the term containing 
the lowest power of x in the quotient. In solving the problem 
in this way, we should first arrange both dividend and divisor 
in order of powers commencing with the lowest, and should 



o 



G ALGEBRA. 



afterwards bring clown each remainder in the same order; re- 
membering that a term which does not contain x at all con- 
tains a lower power of x than any term which contains x. 

101. From the preceding principles we derive the follow- 



ing 



RULE. 

Arrange both dividend and divisor in the same order of pow- 
ers of some common letter. 

Divide the first term of the dividend by the first term of the 
divisor, and write the result as the first term of the quotient. 

Multiply the whole divisor by this term, and subtract the 
product from the dividend, arranging the result in the same 
order of powers as the divisor and dividend. 

Regard the remainder as a new dividend, and divide its first 
term by the first term of the divisor, giving the next term of the 
quotient. 

Multiply the whole divisor by this term, and subtract the 
product from the last remainder. 

Continue in the same manner until the remainder becomes 
zero, or until the first term of the remainder will not contain 
the first term of the divisor. 

When a remainder is found whose first term will not con- 
tain the first term of the divisor, the remainder may be written 
with the divisor under it in the form of a fraction, and added 
to the quotient. 

2. Divide a 8 - 3 a 2 b +. 12 b z + 5 a b 2 by b + a. 
Arranging the dividend and divisor in order of powers, 

a + b) a 3 - 3 a 2 b + 5 a b 2 + 12 b z {a 2 - 4 a b + 9 b 2 
a 3 + a 2 b 

- 4 a 2 b ~ 

— 4 a 2 b — 4 a b 2 

9ab 2 

9 a b 2 +9b s 



3 b 3 , Remainder. 

3 b 9 



Ans, a 2 -4ab + 9b 2 + 



a + b' 



DIVISION. 37 

EXAMPLES. 

3. Divide 2 a 2 x 2 — 5 a x + 2 by 2 a x — 1. 

4. Divide 3 6 3 + 3 a b 2 - 4 a 2 b - 4 « 3 by a + b. 

5. Divide 8 a? - 4 a 2 i - 6 a b 2 + 3 & 3 by 2 a - J. 

6. Divide 21 a 5 - 21 b 5 by la — lb. 

7. Divide a 3 + 2 a; 3 by a + #. 

8. Divide x* + y 4 by cc + y. 

9. Divide 23 x"- 48 + 6 cc 4 - 2 a; - 31 x* by 6 + 3a; 2 — 5 x. 

10. Divide 15 x 4 - 32 x 3 + 50 x 2 - 32 a; + 15 by 3 x 2 - 4 x + 5. 

11. Divide 2 a; 4 - 11 aj - 4 ar - 12 - 3 x 3 by 4 + 2 ar + jb. 

12. Divide a; 5 — v/ 5 by x — y. 

13. Divide 35 - 17 x + 16 x 2 - 25 a; 3 + 6 x 4 by 2 x — 1. 

14. Divide 3 x 2 + 4 x + 6 a; - 11 x 3 - 4 by 3 x 2 - 4. 

15. Divide a 2 — &' 2 + 2 & c — c 2 by a + b — c. 

16. Divide a; 4 — 9 a; 2 — 6 x y — y 2 by x 2 + 3 x + y. 

17. Divide x n + 1 + x n y + x y n + y n + 1 by x n + y n . 

18. Divide cr n — b 2m + 2 b m c r — c 2r by a n + b m — c r . 

19. Divide 1 + a hy 1 — a. 

In examples of this kind the division does not terminate, 
there being a remainder however far the operation may be 
carried. 

20. Divide a by 1 + x. 

21. Divide a 8 + a 6 b 2 + a 4 b 4 + a 2 b* + b a 

by a 4 +a 3 b + a 2 b 2 + ab 3 + b 4 . 

22. Divide 3 a 3 + 2 - 4 a 5 + 7 a + 2 a 6 - 5 a 4 + 10 a 2 

by a 3 — 1 — a 2 — 2 a. 

23. Divide 15 x 2 - x 4 - 20 - 2 a; 5 + 6 x + 2 x 3 

by 5 - 3 a; 2 - 4 x + 2 x 3 . 



38 ALGEBRA. 

24. Divide 2 x 5 + 4 x 2 — 14 + 7 x — 7 x 3 + x« — x i 

by 2 x 2 - 7 + z 3 . 

25. Divide 12 « 5 - 14 a 4 b + 10 a 8 i 2 - a 2 6 3 - 8 a i 4 + 4 i 5 

by 6 a 8 - 4 a 2 6 - 3 a b 2 + 2 6 3 . 

102. In Art. 88 we showed that X a = 0. Since the 
product of the divisor and quotient equals the dividend, we 
may regard as the quotient, a as the divisor, and as the 
dividend. Therefore, 

°- = 0. 

That is, a 

If zero be divided by any quantity the quotient is equal to 
zero. 



VII. — FORMULAE. 

103. A Formula is an algebraic expression of a general rule. 
The following formulae will be found very useful in abridg- 
ing algebraic operations. 

104. By Art. 17, (a + bf =5 (a + b) (a + b) ; whence, by 
actual multiplication, we have 

That is, ( a + V* = cc 2 + 2ab + b 2 . (1) 

The square of the sum of two quantities is equal to the 
square of the first, plus twice the product of the first by the 
second, plus the square of the second. 

105. We may also show, by multiplication, that 

(a - b) 2 = a 2 -2ab + b\ (2) 

That is, 

The square of the difference of two quantities is equal to 

the square of the first, minus twice the product of the first by 

the second, plus the square of the second. 

106. Again, by multiplication, we have 

(a + b) (a-b) = a 2 - b 2 . (3) 



FORMULAE. 3y 

That is, 

The product of the sum and difference of two quantities is 
equal to the difference of their squares. 

EXAMPLES. 

1 107. 1. Square 3 a + 2 b. 

The square of the first term is 9 a 2 , twice the product of the 
terms is 12 a b, and the square of the last term is 4 b 2 . Hence, 
by formula (1), 

(3 a + 2 b) 2 = 9 a 2 + 12 a b + 4 b 2 , Am. 
Square the following : 

2. 2m + 3 w. 4. 3 x + 11. 6. 2ab + &ac. 

3. x 2 + 4. 5. 4a + 5 b. 7. 7 x 3 +3x. 
8. Square 4 x — 5. 

The square of the first term is 16 x 2 , twice the product of 
the terms is 40 a-, and the square of the last term is 25. 
Hence, by formula (2), 

(4 x - 5) 2 = 16 x 2 - 40 x + 25, Am. 

Square the following : 

9. 3a 2 -b s . 11. l-2£ 13. 3-a\ 

10. 4 a b - x. 12. x* - y\ 14. 2 a; 3 - 9 x 2 . 

15. Multiply 6 a + b by 6 « — 6. 

The square, of the first term is 36 a 2 , and of the last term b 2 . 
Hence, by formula (3), 

(6a + b) (6 a - b) = 36 a 2 - b 2 , Am. 
Expand the following : 

16. (x + 3) (x - 3). 19. (a m + a") (a m - a n ). 

17. (2 x + 1) (2 x - 1). 20. O 3 + 5 a-) (a; 3 - 5 a-). 

18. (5« + 7J) (5a-76). 21. (4ar + 3) (4a; 2 -3). 
22. Multiply a + b + chja + b — c. 

(a + b + c) (a + b - c) = [(» + b) + c] [O + 6) - c] 
= (Art. 106), (a + b) 2 - c 2 = a 2 + 2 « b + b 2 - c 2 , Am. 



40 ALGEBRA. 

Expand the following : 

23. [1 + («-&)] [1- («- &)]• 25 - (a-5 + c)(a — &-c). 

24. (a + & + c) (a — & — c). 26. (c + a - &) (c — a + b). 

27. [(a + b) + (c-d)2 [(a + ft) - («-«*)]. 

28. (a — b + c — d)(a — b — c + d). 

29. (a + b + c + d) (a + b — c — d). 



VIII. — FACTORING. 

108. The Factors of a quantity are such quantities as will 
divide it without a remainder. 

109. Factoring is the process of resolving a quantity into 
its factors. 

110. A Prime Quantity is one that cannot be divided, 
without a remainder, by any integral quantity, except itself 
or unity. 

Thus, a, b, and a + c are prime quantities. 
Quantities are said to he prime to each other when they have 
no common integral divisor except unity. 

111. One quantity is said to be divisible by another when 
the latter will divide the former without a remainder. 

Thus, a b and a b + a 2 b' 1 are both divisible by a, b, and a b. 

112. If a quantity can be resolved into two equal factors, 
it is said to be a, perfect square ; and one of the equal factors 
is called the square root of the quantity. 

If a quantity can be resolved into three equal factors, it is 
said to be a perfect cube ; and one of the equal factors is called 
the cube root of the quantity. 

Thus, since 1 a- equals 2 a X 2 a, 4 a 2 is a perfect square 
and 2 a is its square root ; and since 27 X s equals 3,xx3a , x3/, 
27 X s is a perfect cube, and 3x is its <nbe root. 



FACTORING. 41 

Note. 4 a 2 also equals - 2 a x - 2 a, so that the square root of 4 <> 2 
may be either 2 a or - 2 «. In the examples in this chapter we shall only 
consider the positive square root. 

To find the square root of an algebraic quantity, extract the square root 
of the numerical coefficient, and divide the exponent of each letter by 2. 
Thus, the square root of 9 a 6 b 2 is 3 a? b. 

To find the cube root, extract the cube root of the numerical coefficient, 
and divide the exponent of each letter by 3. Thus, the cube root of 8 a 3 b® 
is 2 a b 2 . 

113. The factoring of monomials may be performed by- 
inspection ; thus, 

12 a 3 b 2 c = 2.2.8. a a abbe. 
But in the decomposition of polynomials we are governed by 
rules which may be derived from the laws of their formation. 
A polynomial is not always factorable ; but in numerous cases 
we can always factor ; and these cases, together with the rules 
for their solution, will be found in the succeeding articles. 

CASE I. 

114. Wlien the terms of a polynomial have a common mo- 
nomial factor, it may be written as one of the factors of the 
polynomial, ivith the quotient obtained by dividing the given 
polynomial by this factor, as the other. 

1. Factor the expression 3 x 3 y 2 — 12 x y 4 — 9 x 2 y 3 . 
We observe that each term contains the factor 3 x y 2 . 
Dividing the given polynomial by 3 x y 2 , we obtain as a 
quotient x 2 — 4 y 2 — 3 x y. Hence, 

3 x 3 y 2 - 12 x y 4 - 9 x 2 y z = 3 .« y 2 (x 2 -±y 2 -3x y), Arts. 

EXAMPLES. 

Factor the following expressions : 

2. a s + a. $ 5. 60m 4 n 2 — 12 n\ 

3. 16 x 4 - 12 x. 6. 27 c 4 d 2 + 9 c 3 d. 

4. a & -2 a 4 + 3 a 3 -a 2 . 7. 36 X s y — 60 x 2 y 4 — 84 x 4 y 2 . 

8. a 5 b-3a 6 b 4 -2a 3 b 4 c + 6a' 1 b 5 x. 

9. 84 x 2 y 3 - 140 x 3 y 4 + 56 x 4 if. 



42 ALGEBRA. 

10. 72 o 4 b 3 c 3 + 126 a 3 c 2 d + 162 a 2 c. 

1 1. 128 c 4 d 5 + 320 c 2 d 7 - 448 c 8 <Z 4 . 

CASE II. 

115. TFAerc a polynomial consists of four terms, the first 
two and last two of which have a common binomial factor, it 

may he written as one of the factors of the polynomial, wit li- 
the quotient obtained by dividing the given polynomial by this 
factor, as the other. 

1. Factor the expression a m — b m + a • n — b n. 
Factoring the first two and last two terms by the method 

of Case I, we obtain m (a — b) + n (a- — b). We observe that 
the first two and last two terms have the common binomial 
factor a — b. Dividing the expression by this, we obtain as a 
quotient m + n. 

Hence, am — bm + an — bn=(a — b) (m + n), Ans. 

2. Factor the expression a m — bm — an + b n. 

am — b 7)i — a n + b n =■ a m — b m — (a n — b 11) = ni (a — b) 
— n(a — b) — (a — b) (m — n), Ans. 

Note. If the third term of the four is negative, as in Ex. 2, it is 
convenient to enclose the last two terms in a parenthesis with a - sign 
prefixed, before factoring. 

EXAMPLES. 

Factor the following expressions : 

3. a b + b x + a y + x y. -7. mx 2 — my 2 — ?ix 2 + n y 1 . 

4. a c — cm + a d — dm. 8. x 3 + x 2 + x + 1. 

5. x 2 + 2x — xy — 2y. 9. 6 x 3 + 4 x 2 — 9 x - 6. 

6. a 3 — a 2 b + a b 2 - b 3 . 10. 8 c x - 12 c y + 2 d x - 3 d y. 

11. 6 n - 21 m 2 n-8m + 28 m 3 . 

12. a 2 bc — ac 2 d+ab 2 d — bc d 2 . 

13. m 2 n 2 x 2 — n s x y — m 3 x y + m n y' 2 . 

14. 12 a b m n — 21 a b x y + 20 <■ d m n — 35 c d x //. 



FACTORING. 4:3 

CASE III. 

116. When the first and last terms of a trinomial are 
perfect squares and positive, and the second term is twice the 
product of their square roots. 

Comparing with Formulae 1 and 2, Arts. 104 and 105, we 
observe that such expressions are produced by the product of 
two equal binomial factors. Reversing the rules of Arts. 104 
and 105, we have the following rule for obtaining one of the 
equal factors : 

Extract the square roots of the first and last terms, and 
connect the results by the sign of the second term. 

1. Factor a 2 + 2 a b + b 2 . 

The square root of the first term is a ; of the last term, b ; 
the sign of the second term is + . Hence, one of the equal 
factors is a + b. 

Therefore, a 2 + 2 a b + b 2 = (a + b) (a + b) or (a + b) 2 , Ans. 

2. Factor 9 or — 12 a, b + A hi 

The square root of the first term is 3 a ; of the last term, 2 b ; 
the sign of the second term is — . Hence, one of the equal 
factors' is 3 a — 2 b. Therefore, 

9 a 2 - 12 a b + 4 b 2 = (3 a -2 b) (3 a - 2 b) or (3 a - 2 bf, Ans. 

Note. According to Art. 58, the given expression may be written 
4 b 2 — 12 a b + 9 a 2 . Applying the rule to this expression, we have 

4 b 2 - 12 a b + 9 a 2 = (2 b- 3 a) (2 b - 3 a) or (2 b - 3 a) 2 . 

We should obtain this second form of the result in another way by apply- 
ing the principles of Art. 89 to the first factors obtained. 

EXAMPLES. 

Factor the following expressions : 

3. x 2 - 14 x + 49. 6. a 2 - 28 a + 196.. 

4. m 2 + 36 m + 324. 7. n 6 - 26 n 3 + 169. 

5. y 2 + 20 y + 100. 8. x 2 y 2 + 32 x y + 256. 



44 ALGEBRA. 

9. 25 x 2 + 70 x y z + 49 y 2 z 2 . 11. 16 m 2 -8am+« 2 . 

10. 36 m? - 36 w « + 9 n 2 . 12. 4 a 2 + 44 a b + 121 b 2 . 

13. a 2 6 4 + 12 a b 2 c + 36 c 2 . 

14. 9 « 4 + 60 a 2 bc 2 d + 100 b 2 c i d 2 . 

15. 4 x A - 60 m re x 2 + 225 m 2 t»s 

16. 64 x 6 - 160 x 5 + 100 cc 4 . 

CASE IV. 

117. When an expression is the difference between two 
perfect squares. 

Comparing with Formula 3, Art. 106, we observe that such 
expressions are the product of the sum and difference of two 
quantities. Reversing the rule of Art. 106, we have the fol- 
lowing rule for obtaining the factors : 

Extract the square roots of the first and last terms ; add 
the results for one factor, and subtract the second result from 
the first for the other. • 

1. Factor 36 x 2 — 49 y 2 . 

The square root of the first term is 6 a;; of the last, 7 y. 
The sum of these is 6 x + 7 y, and the second subtracted from 
the first is 6 x — 7 y. Hence, 

36 x 2 — 49 y 2 = (6 x + 7 y) (6 x — 7 y), Ans. 

2. Factor {a - b) 2 - (c-df. 

The square root of the first term is a — b ; of the last, c — d. 
The sum of these is a — b + c — d, and the second subtracted 
from the first is a — b — c + d. Hence, 

(a — b) 2 — (c — d) 2 = (a — b + c — d) (a — b — c + cl), Ans. 

EXAMPLES. 

Factor the following expressions : 

3. x 2 -l. 5. a 4 -/A 7. 4 a* — 225 m a w 9 . 

4. 4x 2 -9t/ 2 . 6. 9 a 2 -4. 8. 1 - 196 x 2 if z\ 



FACTORING. 45 

9. (a + b) 2 - (c + d) 2 . 11. m 2 - (x - y) 2 . 

10. (a-c)*—b*. 12. (aj — m) a — (y — »)*. 

Many polynomials, consisting of four or six terms, may be 
expressed as the difference between two perfect squares, and 
hence may be factored by the rule of Case IV. 

13. Factor 2 m n + m 2 — 1 + n 2 . 

Arrange the expression as follows, m 2 + 2mn + r? — 1. 
Applying the method of .Case III to the first three terms, we 
may write the expression (m + n)' 2 — 1. The square root of 
the first term is m + n ; of the last, 1. The sum of these is 
m + n + 1, and the second subtracted from the first is 
m + n — 1. Hence, 

2 m n + m 2 — 1 + n 2 = (m + n + 1) (m + n — 1), Ans. 

14. Factor 2 x y + 1 — x 2 — y 2 . 

2 x y + 1 — x 2 — y 2 = l — x 2 -\-2x y — y 2 
= 1- {x 2 -2 x y + y 2 ) = l - (x-y) 2 , by Case III, 
= [1 + (x— y)][l — (as— y)] = (l + #-y) (1— x + y),Ans. 

15. Factor 2xy+b 2 -x 2 -2ab- y 2 + a 2 . 

2xy+b 2 — x 2 — 2ab — y 2 + a 2 
= a 2 — 2ab + b' 2 — x 2 + 2 x y — y 2 
= a 2 -2 a b + b 2 - (x s - 2 x y + y 2 ) 
= (a - b) 2 — (x— y) 2 , by Case III, 
= [(«-&)+ (as-y)][(a-fl)-(aj-y)] 
= (a — b+x — y) (a—b — x + y), Ans. 

Factor the following expressions : 

16. x 2 + 2xy + y 2 -4:. 19. 9-x 4 -4?/ 2 + 4f 2 y. 

17. a a _ j2 + 2 h c _ c 2. 20. 4 a 2 + 6 2 - 9 d? -4ab. 

18. 9 c 2 - 1 + d? + 6 c d. 21. 4 b - 1 - 4 b 2 + 4 m 4 . 

22. a 2 — 2 a m + m 2 — b 2 — 2 b n — n 2 . 

23. 2 a m — 6 2 + m 2 + 2bn+ a 2 - n 2 . 

24. x 2 - y 2 + c 2 - d 2 - 2 c x + 2 tf y. 



46 ALGEBRA. 



CASE V. 



118. Wlien an expression is a trinomial, of the form 
x- + a x + b ; where the coefficient of x 2 is unity, and a and 
b represent any whole numbers, either positive or negative. 

To derive a rule for this case we will consider four examples 
in Multiplication : 

I. ii. 

x + 5 x — o 

x + 3 x — 3 



x 1 + 5 x x 2 —ox 

+ 3.x- + 15 -3cc + 15 

a; 2 + Sx + 15 sc a -8aT+15 

III. IV. 

x + 5 x — 5 

x — 3 x + 3 



a? 2 + 5 x # 2 — 5 .<■ 

-3x-15 +3x-ll 



« 2 +2^-15 x 2 -2ic-15 

We observe in these results, 

1. The coefficient of x is the algebraic sum of the numbers 
in the factors. 

2. The last term is the product of the numbers. 

Hence, in reversing the process, we have the following rule 
for obtaining the numbers : 



RULE. 

Find two numbers ivhose algebraic sum is the coefficient of 
x, and whose product is the last term. 

Note. We may shorten the work by considering the following points : 

1. When the last term of the product is f, as in Examples I ami II, 
the sum of the numbers is the coefficient of .<• ; both numbers being + 
when the second term is +, and - when the second term is -. 

2. When the last term is -, as in Examples III and I V, the difference 



FACTORING. 47 

of the numbers (disregarding signs) is the coefficient of x ; the larger 
number being + and the smaller - when the second term is +, and the 
larger number - and the smaller + when the second term is - . 

We may embody these observations in the form of a rule which may be 
found more convenient than the preceding rule in the solution of examples. 

I. If the last term is +, find tivo members whose sum is the coefficient of 
x, and whose product is the last term; and give to both numbers the sign of 
the second term. 

II. If the last term is - , find tivo numbers whose difference is the coeffi- 
cient of x, and whose product is the last term; and give to the larger num- 
ber the sign of the second term, and to the smaller number tlie opposite sign. 

1. Factor x 2 + 14 x + 45. 

Here we are to find two numbers whose - " - 

(.product — 45 J 

The numbers are 9 and 5 ; and, the second term being + , both 
numbers are +. Hence, 

x 2 + 14 x + 45 = (x + 9) (x + 5), Ans. 

2. Factor x 2 — 6 x + 5. 

Here we are to find two numbers whose ■! \ 

(product = 5,' 

The numbers are 5 and 1 ; and, as the second term is — , both 
numbers are — . Hence, 

x 2 — 6 x + 5 = (x — 5) (x — 1), Ans. 

3. Factor x 2 + 5 x — 14. 

Here we are to find two numbers whose ■< " 

I product = 14 ) 

The numbers are 7 and 2; and as the second term is +, the 
larger number is + , and the smaller — . Hence, 

x 2 + 5 x — 14 = (x + 7) (x — 2), Ans. 

4. Factor x 2 — 7 x — 30. 

Here we are to find two numbers whose ] ~ \ 



48 ALGEBKA. 

The numbers are 10 and 3; and as the second term is — , the 
larger number is — , and the smaller + . Hence, 

x 2 - 7 x - 30 = (x - 10) (x + 3), Arts. 

Note. In case the numbers cannot be readily determined by inspection, 
the following method will always give them : 

Eequired two numbers whose difference is 8 and product 48. Taking in 

order, beginning with the lowest, all possible pairs of integral factors of 48, 

we have 

1x48, 

2x24, 

3x16, 

4x12. 

And, as 4 and 1 2 differ by 8, they are the numbers required. 

Evidently this method will give the required numbers eventually, how- 
ever large they may be, provided they exist. 

EXAMPLES. 

Factor the following expressions : 

5. a 2 +5 a- + 6. 12. m 2 +9m + 8. 

6. a 2 -3a + 2. 13. m' 2 + 2m-80. 

7. 2/2 + 2?/- 8. 14. c 2 - 18 c + 32. 

8. m 2 -5m-24. 15. x 2 + x-42. 

9. * 2 -ll;c + l8. 16. x 2 + 23x + 102. 

10. n *- n -c)0. 17. if-9y-90. 

11. * 2 +13a; + 36. 18. a 2 +13a-48. 

19. cc 2 -9z-70. 

20. Factor 15 — 2x — x 2 . 

15 _ 2 x - x 2 = - (x 2 + 2 x - 15) 

By the rule of Case V, x 2 + 2 x - 15 = (.r + 5) (x - 3). 

Hence, 

15 - 2 x - x 2 = - (x + 5) (x - 3) = (x + 5) (3 - x), Ans. 

Note. If the x* term is -, enclose the whole expression in a paren- 
thesis with a - sign prefixed. Factor the quantity within the parenthesis, 
and change the signs of all the terms of one factor. 



FACTORING. 49 

Factor the following expressions : 
21. 20-x-x 2 . 22. 6 + x-x 2 . 23. 84-5z-a; 2 . 

The method of Case V may he extended to the factoring of 
more complicated trinomials. 

24. Factor m 2 ri 2 — 3 m nx + 2 x 2 . 

r. i it (sum =3) 

Here we are to find two numbers whose < t — 9 j 

The numbers are 2 and 1; and as the second term is—, 
hoth numbers are — . Hence, 

m 2 ri 2 — 3 m nx + 2x 2 = (m n — 2 x) (m n — x), Ans. 

Factor the following expressions : 

25. a- 4 - 29 x 2 + 120. 30. ?« 4 + 5 m 2 n 2 - 66 n\ 

26. c 6 + 12c 3 +ll. 31. (a-b) 2 -S(a-b)-4. 

27. x 2 y G + 2xf-120. 32. (x + y)*- 7 (x + y) + 10. 

28. (r '^_7«i 2 -144. 33. x 2 - 2 x y 2 z - 48 y* z\ 

29. x 2 + 25 w x + 100 ?i 2 . 34. (m + nf + (m + n) - 2. 

CASE VI. 

119. When an expression is the sum or difference of two 
perfect cubes. 

By actual division, we may show that 

a 3 + b 3 a 3 — b 3 

— — = a 2 — ab + b' 2 , and — = a 2 + a b + b 2 . 

a + b a — b 

Whence, 

(a 3 + b 3 ) = (a + b) (a 2 -ab + b 2 ), and 

(a 3 - b 3 ) = (a - b) (a 2 + ab + b 2 ). 

These results may he enunciated as follows : 

To factor the sum of two perfect cubes, write for the first 
factor the sum of the cube 7'oots of the quantities; and for the 



50 ALGEBEA. 

second, the square of the first term of the first factor, minus 
the product of the two terms, phis the square of the last term. 
To factor the difference of two perfect cubes, write for the 
first factor the difference of the cube roots of the quantities ; 
and for the second, the square of the first term of the first 
factor, plus the product of the two terms, plus the square of 
the last term. 

1. Factor 8 a 3 + 1. 

The cube root of the first term is 2 a ; of the last term, 1. 
Hence, 8 a 3 + 1 = (2 'a + 1) (4 a 2 - 2 a + 1), Ans. 

2. Factor 27 x 6 - 64 y 3 . 

The cube root of the first term is 3 x 2 ; of the last term, 4 y. 
Hence, 

27 x G - 64 y* = (3 x 2 - 4 y) (9 x i + 12 x 2 y + 16 y 2 ), Ans. 

EXAMPLES. 

Factor the following expressions : 

6. Sc 6 -d 9 . 9. 343 + 8 a 3 . 

7. 125 a s - 216 m 3 . 10. 27 a; 3 -125. 

8. 729 c 3 dP + 512. 11. 1000 -27 a 3 b\ 

CASE VII. 

120. When an expression is the sum or difference of two 
like powers ofttvo quantities. ■ 

The following principles are useful to remember : 

1. a n — b n is always divisible by a — b, if n is an. integer. 

2. a n — b n is always divisible by a + b, if n is an even integer. 

3. a n + b n is always divisible by a + b, if n is an odd integer. 

We may prove the first principle as follows : 
Commencing the division of a n — b n by a — b, wv have 



3. 


iC 3 — y 3 . 


4. 


a 3 + 8. 


5. 


m 3 + 64 ?z, 6 . 



a" — b n 



FACTORING. 51 

a — b 



a n ' + . . . Quotient. 



- 



a n ] b — b n Remainder. 

a n — b n , a n - } b — b n , b (a n ~ l — b n ~ v ) 

or, - — - =a n ~ i H ; = a n ~ l -\ '-, 

a — b a — b a — b 

It is evident from this result that, if a"" 1 — & n_1 is exactly 
divisible by a — b, the dividend a n — b n will be exactly divisi- 
ble by a, — b. That is, if the difference of two like powers of 
two quantities is divisible by the difference of those quantities, 
then the difference of the next higher powers of the same 
quantities is also divisible by the difference of the quantities. 
But a s — b 3 is divisible by a — b, hence a 4 — b 4 is ; and since 
c^ — b* is divisible by a — b, a 5 — b 5 is; and so on to any 
power. This proves the first principle. 

Similarly the second and third principles may be proved. 

By continuing the division, we should find, 

= a"- 1 + a n ~ 2 b + a n ~ 3 b~ + + ab n ~ 2 + b n - 1 (1) 

a n ~' 2 b + a n - s b 2 — + ab n ~ 2 — fi"" 1 ' (2) 



a 


-b 




10 




a n 
a 


— b n 

+ b 


= 


a n - 


-l 


a n 

L. 


+ b n 


— 


a n ~ 


-l 



a + b 



a 



2 b + a n - 3 b 2 - —ab n - 2 +b n ~ 1 (3) 



It is useful to remember that when a — b is the divisor, all 
the terms of the quotient are + ; where a + b is the divisor, 
the terms of the quotient are alternately + and — , the last 
term being + if n is odd, and — if n is even., 

1. Factor a 1 — IP. 
Putting n — 7 in (1), we have 



a 1 -b 



a* + a 5 b + a 4 b 2 + a 3 b 3 + a 2 b* + ab s + b 6 . 



a — b 
Hence, 

a 7 - V = (a - b) (a 6 + a 5 b + a 4 b 2 + a 3 b 3 + a 2 b 4 + a b 5 + b 6 ), 

Ans. 



52 ALGEBRA. 

2. Factor m 5 + x 5 . 

Putting a = m, b = x, n== 5, in (3), we have 

1YI -f- X . „ 2 2 3,4 

= m — m° x + m x — mx A + x\ 

m + x 

Hence, 

m 5 + x s = (m + x) (ra 4 — m 3 a; + ra 2 a: 2 — m x 3 + x 4 ), Ans. 

3. Factor x 6 — y 6 . 

Putting a = x, b = y, n = 6, in (1), we have 

x 6 ifi 

— = x s + x* y + x s y 2 + x 2 y 3 + x y* + y 5 . 

Hence, 

x 6 — y 6 = (x — y) (x 5 + x 4 y + x s y 2 + x 2 y s + x y* + f), Ans. 

Or, putting a =x, b = y, ?i = 6, in (2), we have 
x 6 ?/ 6 

J_ — ^5 ~,i », I ~,3 „fi ™2 „.3 i_ ™ ,.4 „.5 



Hence, 



= x° — a;* ?/ + x" y — x" y 6 + x if — y" 



x 6 — y 6 =(x + y) (x h — x 4 y + x s y 2 — x 2 y 3 + xy i — y s ), Ans. 

EXAMPLES. 

Factor the following expressions : 

4. x 5 + y 5 . 6. to 6 -c 6 . 8. m s -n*. 10. a 4 -16. 

5. (*-d\ 7. a' + V. 9. c 7 -l. 11. a 7 +128. 

121. By the application of one or more of the given rules 
for factoring, a quantity may sometimes be separated into 
more than two factors. 

1. Factor 2 a x s y 2 - 8 a x y\ 

By Case I, 2 a x 3 if - 8 a x if = 2 a x y 2 (x 2 - 4 if). 
Factoring the quantity in the parenthesis by Case IV, 
2 a x 3 if — 8 a x if = 2 a x y 2 (x + 2 y) (x — 2 y), Ans. 

Note. If the method of Case I is to be used in connection with other 
cases, it should be applied first. 



GREATEST COMMON DIVISOR. 53 

2. Resolve a G — b 6 into four factors. 
By Case IV, a 6 - b G = (a 3 + b 3 ) (a 3 - b 3 ). 
By Case VI, a 3 + b 3 = (a + b) (a 2 -ab + b 2 ), 
and a 3 - b 3 = (a - b) (a 2 + ab + b 2 ). 
Hence, 
a 6 -b G =(a + b) (a- b) (a 2 -ab + b 2 ) (a 2 +ab + b 2 ), Ans. 

EXAMPLES. 

Factor the following expressions : 

3. 3 a 3 b + 12 a 2 b + 12 a b. 7. 3 « 4 - 21 a 3 + 30 a: 

4. 45 x 3 if — 120 x 2 f + 80 x if. 8. 2 c 3 m + 8 c 2 ra-42 c m. 

5. 18 x 3 y — 2 x f. 9. m 2 a;y-4wa;y— 12 ay. 

6. x 3 + Sx 2 +7x. 10. 32 « 4 b + 4 a 6 4 . 

11. Resolve n 9 — 1 into three factors. 

12. Resolve x i — if into three factors. 

13. Resolve x 8 — m 8 into four factors. 

14. Resolve m 6 — ?i 6 into four factors. 

15. Resolve a 9 + c 9 into three factors. 

16. Resolve 64 « 6 — 1 into four factors. 

Other methods for factoring will he given in Chapter XXIX. 



IX.— GREATEST COMMON DIVISOR. 

122. A Common Divisor or Measure of two or more quan- 
tities is a quantity that will divide each of them without a 
remainder. 

Hence, any factor common to two or more quantities is a 
common divisor of those quantities. 

Also, when quantities are prime to each other, they have no 
common measure greater than unity. 



54 ALGEBRA. 

123. The Greatest Common Divisor of two or more quan- 
tities is the greatest quantity that will divide each of them 
without a remainder. 

Hence, the greatest common divisor of tivo or more quanti- 
ties is the product of all the prime factors common to those 
quantities. 

By the greatest of two or more algebraic quantities, it may 
he remarked, is here meant the highest, with reference to the 
coefficients and exponents of the same letters. 

In determining the greatest common divisor of algebraic 
quantities, it is convenient to distinguish three cases. 

CASE I. 

124. When the quantities are monomials. 
1. Find the greatest common divisor of 

42 a 3 b 2 , 70 a 2 b c, and 98 a 4 b 3 d 2 . 

42 a 3 b 2 =2x3x7 a a a bb 
70 a 2 be =2x5x7 aa b c 
98 a 4 b 3 d 2 = 2 x 7 X 7 aaaabbb del 



Hence, G. C. D. = 2 X 7 a a b = 14 d 2 b, Ans. (Art. 123). • 

RULE. 

Resolve the quantities into their prime factors, and find the 
product of all the factors common to the several quantities. 

Note. Any literal factor forming a part of the greatest common divisor 
will take the lowest exponent with which it occurs in either of the given 
quantities. 

EXAMPLES. 

Find the greatest common divisors of the following : 

2. a s x 2 , 7 a* x, and 3 a b 2 . 

3. G c 5 d\ 3 c 3 d 5 , and 9 c* d 3 . 

4. 18 m n 5 , 45 m 2 n, and 72 m 8 ri 2 . 

5. 15 c 2 x, 45 c 3 x 2 , and 60 c 4 x 8 . 



GREATEST COMMON DIVISOR. 55 

6. 108 y 2 z\ 144 f z\ and 120 if z 5 . 

7. 96 a 5 b\ 120 a 3 b 5 , and 168 a 4 b & . 

8. 51 m 4 n, 85 ra 3 a-, and 119 m 2 if. 

9. 84 a 8 y 4 z s , 112 a; 4 v/ 5 z 6 , and 154 a 7 y 6 z\ 

CASE II. 

125. When the quantities are polynomials which can be 
readily factored by inspection. 

1. Find the greatest common divisor of 

5xy 3 — 15y 3 , x 2 + 4 x — 21, and mx — 3m — nx + 3n. 

5xy 3 — 15y 3 = 5y 3 (x — 3) 

x 2 + 4 x - 21 = (x + 7) (a; - 3) 
mx — 3 m — nx + 3n = (m — n) (x — 3) 

Hence, by Art. 123, G. C. D. = x — 3, Ans. 

2. Find the greatest common divisor of 

4 x 2 - 4 x + 1, 4 x 2 - 1, and 8 ar 5 - 1. 

4a; 2 -4a; + l = (2a;-l) (2 a; - 1) 
4 a 2 - 1 = (2 x + 1) (2 x-1) 
8 a? - 1 = (2 x - 1) (4 a; 2 + 2 x + 1) 

Hence, G. C. D. = 2 a: — 1, Ans. 

The rule in this case is the same as in Case I. 

EXAMPLES. 

Find the greatest common divisors of the following : 

3. 3 a x 2 — 2 a 2 x, a 2 x 2 — 3 a b x, and 5 a 2 x 3 + 2 ax 4 — 3 a 3 x. 

4. vi 2 + 2 in n + n 2 , m 2 — n 2 , and m 3 + n 3 . 

5. x* — 1, a- 5 + a 3 , and a; 4 + 2 a; 2 + 1. 

6. 3 a xif + 21 ay 2 , 3 c a- + 21 c - 3 d x - 21 rZ, and a; 2 -3a- - 70. 

7. 4 x 2 — 12 a + 9, 4 a; 2 - 9, and 4 m 2 »a;-6 m 2 n. 

8. 9a 2 -16, 3a;y — 4^ + 3a;2 — 4a, and 27 a; 3 — 64. 



56 ALGEBRA. 

9. x s — x, X s + 9 x 2 — 10 x, and x 6 — x. 

10. a 3 — 8 ft 3 , 5 a x + 2 a — 10 ft x — 4 ft, and a 2 — 4 a ft + 4 ft 2 . 

11. ar 2 - a; - 42, x 2 - 4 a - 60, and x 2 + 12 x + 36. 

12. 8 x 3 + 125, 4 aj 2 - 25, and 4 a 2 + 20 x + 25. 

13. 3 a x G — 3 a x 5 , a x s — 9 a x 2 + 8 a x, and 2 a x h — 2 a x. 

14. 12 a x - 3 a + 8 c x - 2 c, 64 x 3 - 1, and 16 as* — 8 a? + 1. 

CASE III. 

126. When the quantities are polynomials which cannot be 
readily factored by inspection. 

Let a and ft be two expressions, arranged in order of powers 
of some common letter ; and let the exponent of the highest 
power of that letter in ft be either equal to or less than the 
exponent of the highest power of that letter in a. Suppose 
that ft is contained in a, p times with a remainder c ; suppose 
that c is contained in ft, q times with a remainder d ; and 
suppose that d is contained in c, r times with no remainder. 
The operation of division may be shown as follows : 

ft) a (p 
p ft 

e) b (q 
1 c 



d) c (r 
rd 


We will first show that d is a common divisor of a and ft. 
From the nature of subtraction, the minuend equals the sub- 
trahend plus the remainder ; hence, 

a=pb + c, ft = q c + d, and c = rd. 

Substituting r d for c in tho value of ft, we have 

b = q r d + d = d (q r + 1). 



GREATEST COMMON DIVISOR. 57 

Substituting q r d + d for b, and r d for c in the value of a, 
we have 

a=p q r d + p d + r d = d (p q r + p + r). 

Hence, as d is a factor of a and also of b, it is a common 
divisor of a and b. 

We will now show that every common divisor of a and b is 

a divisor of d. Let k he any common divisor of a and b, such 

that fl^ffli and b = n k. From the nature of subtraction, 

the minuend minus the subtrahend equals the remainder ; 

hence, 

c = a — p b, and d = b — q c. 

Substituting m k for a, and n k for b in the value of c, we 

have 

c = m k — p n k. 

Substituting mk — pnk for c, and n k for b in the value of 
d, we have 

d — nk — q (j)ik—pnk) = nk — qmk+pqnk 
= k (n — q m + p q n). 

Hence, k is a factor or divisor of d. 

Therefore, since every common divisor of a and b is a divisor 
of d, and no expression greater (Art. 123) than d can be a 
divisor of d, it follows that d is the greatest common divisor 
of a and b. 

1. Find the greatest common divisor of x 1 — 6 x + 8 and 
4 x 3 - 21 x 2 + 15 x + 20. 

x 2 - 6 x + 8) 4 a 3 - 21 x 2 + 15 x + 20 (4 x + 3 

4 x 3 - 24 a; 2 + 32 x 

3 x 2 - 17 x + 20 
3 a; 2 - 18 a; + 24 

x _ 4) ^ — 6x + 8(x-2 
x 2 — 4aj 
. -2a; + 8 
-2a + 8 

Hence, cc — 4 is the greatest common divisor, Ans. 



58 ALGEBRA. 

RULE. 

Divide the greater quantity (Art. 123) by the less • and if 
there is no remainder, the less quantity trill he the required 
greatest common divisor. 

If there is a remainder, divide the divisor by it, and continue 
thus to make the preceding divisor the dividend, and the re- 
mainder the divisor, until a divisor is obtained which leaves no 
remainder ; the last divisor will be the greatest common divisor 
required. 

Nota 1. If there are three or more quantities, find the greatest common 
divisor of two of them ; then of this result and the third of the quantities, 
and so on. The last divisor will be the greatest common divisor required. 

Note 2. If a monomial factor is seen by inspection to be common to all 
the terms of one of the given quantities, and not of the other, it may be re- 
moved, as it evidently can form no part of the greatest common divisor ; 
and, similarly, we may remove from a remainder any monomial factor 
which is not a common factor of the given quantities. 

2. Find the greatest common divisor of 

6 a x~ — 19 a x + 10 a and 6 x s — x 2 — 35 x. 

In the first quantity a is a common factor of all the terms, 
and is not a factor of the second quantity ; in the second quan- 
tity x is a common factor of all the terms, and is not a factor 
of the first quantity. Hence we may remove a from each 
term of the first quantity, and x from each term of the second. 

6 a; 2 - 19 a + 10)6 cc 2 - x -35(1 

6a: 2 -- 19:c + 10 
18a;-45 

In this remainder 9 is a common factor of all the terms, and 
is not a common factor of the given quantities. Hence 9 may 
be removed from each term of the remainder. 

2 x - 5)6 x 1 - 19 x + 10(3 x - 2 
6 x 2 — 15 x 

— 4 x'+ 10 

— 4 a; + 10 

Hence, 2 x — 5 is the greatest common divisor, Ans. 



GREATEST COMMON DIVISOR. 59 

Note 3. If the first term of a remainder be negative, the sign of each 
term may be changed. 

3. Find the greatest common divisor of 2 x 2 — 3 a; — 2 and 
2a; 2 -5a;-3. 

2a; 2 - 3 x - 2)2 a 2 - 5 x -3(1 
2x 2 -3x-2 



-2x-l 

The first term of this remainder being negative, we change 
the sign of each term, giving 2 x + 1. 

2 x + 1)2 a;' 2 - 3 x - 2 (x — 2 
2 x' 2 + x 



— 4x — 2 

— 4a; — 2 

Hence, 2 x + 1 is the greatest common divisor, Ans. 

Note 4. The dividend or any remainder may be multiplied by any 
quantity which is not a common factor of all the terms of the divisor. 

4. Find the greatest common divisor of 2 a- 3 — 7 xr + 5 x — 6 
and 3 a' 3 — 7 a' 2 — 7 x + 3. 

To avoid a fraction as the first term of the quotient, we 
multiply each term of the second quantity by 2, giving 
6 X s - 14 x 2 - 14 x + 6. 

2 x s — 7 x 2 + 5 x — 6)6 x s - 14 cc 2 - 14 x + 6 (3 

6 a; 3 - 21 x 2 + 15 a; - 18 



7 a- 2 -29 a; + 24 



To avoid a fraction as the first term of the next quotient, 
we multiply each term of the new dividend by 7, giving 
14 x 3 - 49 x 2 + 35 x - 42. 

7 a; 2 - 29 x + 24) 14 x 3 - 49 x 2 + 35 x - 42 (2 a; 
14 a; 3 - 58 a 2 + 48 x 

9 a- 2 - 13 x - 42 



60 ALGEBRA. 

The first term of this remainder not heing exactly divisible 
by the first term of the divisor, we multiply each term hy 7, 
giving 63 x 2 — 91 x — 294. 

7 x 2 - 29 x + 24) 63 x 2 - 91^-294(9 
63 x 2 - 261 x + 216 



170 x - 510 

Dividing each term by 170, x — 3) 7 x 2 — 29 x + 24 (7 x — 8 

7 x 2 -21x 



— 8 a; + 24 

- 8cc + 24 

Hence, x — 3 is the greatest common divisor, Ans. 

Note 5. When the two given quantities have a common monomial 
factor, it may be removed from each, and the greatest common divisor of 
the resulting expressions found. This result must be multiplied by the 
common monomial factor to give the greatest common divisor of the given 
quantities. 

5. Find the greatest common divisor of 6 x 3 — x 2 — 5 x and 

21 x 3 - 26 x 2 + 5x. 

The quantities have the common monomial factor x ; remov- 
ing it, we find the greatest common divisor of 6 x 2 — x — 5 and 
21 x 2 — 26 x + 5. We multiply the latter by 2, to avoid a frac- 
tion as the first term of the quotient, giving 42 a; 2 — 52 x + 10. 

6 x 2 - x - 5) 42 x 2 - 52 x + 10 ( 7 
42 x 2 - 7a;-35 



— 45 x + 45 



Dividing by — 45, x — 1)6 x 2 — x — 5(6a: + 5 

6 x 2 — 6 x 



5x — 5 
5x — 5 



LEAST COMMON MULTIPLE. 61 

Hence, x — 1 is the greatest common divisor of 6 x 2 — x — 5 
and 21 x 2 — 26 x + 5. Multiplying by x, the common mo- 
nomial factor, we obtain x (x — 1) or x 2 — x as the required 
greatest common divisor, Ans. 

EXAMPLES. 

Find the greatest common divisors of the following : 

6. 6 x 2 — 1 x — 24 and 12 x 2 + 8 x - 15. 

7. 24 x 2 + 11 x - 28" and 40 x 2 - 51 x + 14. 

8. 2 x s -2 x 2 - 3 x + 3 and 2 x 3 -2 x 2 - ox + 5. 

9. 6 x 2 - 13 x -28 and 15 x 2 + 23 a; + 4. 

10. 8 x 2 - 22 x + 5 and 6 a; 2 - 23 x + 20. 

11. 5 x 2 + 58 jc + 33 and 10 .x 2 + 41 x + 21. 

12. x 3 + 2 x 2 + x + 2 and x i -4r-x-2. 

13. 2a; 3 -3cc 2 -;c + l and 6x s — x 2 + 3x -2. 

14. a,- 4 - x 3 + 2 x 2 + x + 3 and x i + 2 x 3 — x — 2. 

15. ft 2 -5ax + 4 x 2 and a 3 — « 2 x + 3 a x 2 — 3 x 3 . 

16. x A -x 3 -5 x 2 + 2 a- + 6 and x 4 + x 3 - x 2 -2x- 2. 

17. 6 x 2 y + 4 x y 2 — 2 y 3 and 4 cc 3 + 2 x 2 ?/ — 2 a* //\ 

18. 2« 4 + 3a 3 x-9a 2 x 2 and 6 « 3 - 17 a 2 ;c + 14 a ar- 3x 3 . 

19. 15 a 2 x 3 - 20 a 2 x 2 - G5 a 2 x - 30 a 2 and 12 6 a; 3 + 20 b x 2 

— 16 b x — 16 b. 



X. — LEAST COMMON MULTIPLE. 

127. A Multiple of a quantity is any quantity that can be 
divided by it without a remainder. 

Hence, a multiple of a quantity must contain all the prime 
factors of that quantity. 



62 ALGEBRA. 

128. A Common Multiple of two or more quantities is one 
that can be divided by each of them without a remainder. 

Hence, a common multiple of two or more quantities must 
contain all the prime factors of each of the quantities. 

129. The Least Common Multiple of two or more quanti- 
ties is the least quantity that can be divided by each of them 
without a remainder. 

Hence, the least common multiple of two or more quantities 
must be the pmxluct of all their different prime factors, each 
taken only the greatest number of times it is found in any one 
of those quantities. 

By the least quantity, is here meant the lowest with refer- 
ence to the exponents and coefficients of the same letters. 

In determining the least common multiple of algebraic 
quantities, we may distinguish three cases. 

CASE I. 

130. When the quantities are monomials. 

1. Find the least common multiple of 36 a s x, 60 a 2 y 2 , and 
84 c x s . 

36 a 3 # = 2x2x3x3 a a ax 
60aV=2x2x3x5 a a yy 

84 c x z =2x2x3x7 x x x c 

Hence, L. C. M. = 2 x f 2 X3x3x5 X? aaaxxxyyc 
= 1260 a 3 x s if c, Ans. (Art. 129). 

RULE. 

Resolve the quantities into their prime factors; and the 
product of these, taking each factor only the greatest number of 
times it enters into any one of the quantities, will be the least 
common multiple. 

Any literal factor forming a part of the least common mul- 
tiple will take the highest exponent with which it occurs in 
p.ither of the given quantities. 



LEAST COMMON MULTIPLE. 6 



o 



When quantities are prime to each other, their product is 
their least common multiple. 

EXAMPLES. 

Find the least common multiples of the following : 

2. 8 a 4 c, 10 a 3 b, and 12 a 2 b 2 . 

3. 5 x 3 y, 10 if z, and 15 x z 3 . 

4. a 5 b 2 , 9 a 3 b\ and 12 a 2 b 3 . 

5. 24 m 3 x 2 , 30 n 2 y, and 32 x y 2 . 

6. 8 c 2 d 3 , 10 a e, and 42 a 2 d. 

7. 36 x y 2 z 3 , 63 x 3 y z 2 , and 28 .r 2 y 3 z. 

8. 40 a 2 b d 3 , 18 a c 3 d\ and 54 J 2 c d\ 

9. 7 m w 2 , 8 x 3 y 2 , and 84 n x y 3 . 

CASE II. 

131. When the quantities are polynomials which can be 
readily factored by inspection. 

1. Find the least common multiple of x 2 + x — 6, x 2 — 6 x + 8 
and x 2 — 9. 

x * + a, _ 6 = (x - 2) (x + 3) 

a;2_6a: + 8 = (a;-2) (cc-4) 
x 2 -9 =(x-3)(x + 3) 

Hence (Art. 129), L. C. M. = (x - 2) (a; - 3) (sc + 3) (x - 4) 
or, x 4 - 6 x 3 - x 2 + 54 x - 72, ^ws. 

The rule is the same as in Case I. 

EXAMPLES. 

Find the least common multiples of the following : 

2. a x 2 + a 2 x, x 2 — a 2 , and x 3 — a 3 . 

3. 2 a 2 + 2 a b , 3 a b - 3 & 2 , and 4 a 2 c — 4 ft 2 c. 



64 ALGEBRA. 

4. x 2 + x, x z — x, and x i + x. 

5. 2 - 2 a- 2 , 4 - 4 a-, 8 + 8 a, and 12 + 12 a 2 . 

6. x 2 + 5x + 4, x 2 + 2x — S, and a; 2 + 7 jc + 12. 

7. x 3 — 10 a; 2 + 21 a;, and a x 2 + 5 a x — 24 a. 

8. 4 ar - 4 a; + 1, 4 x 2 - 1, and 8 a 3 - 1. 

9. a x — a y — b x + b y, x 2 — 2 x y + y 2 , and 3 arb — 3ab 2 . 

10. 9 a; 2 + 12 a; + 4, 27 x 3 + 8, and 6 a x 3 + 4 a x 2 . 

11. cc 2 - 4 cc + 3, x 2 + a; - 12, and x 2 -x- 20. 

12. x 2 — y 2 — z 2 + 2 y z and x 2 -i/ 2 +r + 2a;s. 

CASE III. 

132. When the quantities are polynomials which cannot be 
readily factored by inspection. 

Let a and b be two expressions ; let d be their greatest com- 
mon divisor, and m their least common multiple. Suppose 
that d is contained in a, x times, and in b, y times ; then, from 
the nature of the greatest common divisor, x and y are prime 
to each other. Since the dividend is the product of the quo- 
tient and divisor, we have 

a = dx and b = d y. 

Then (Art. 129) the least common multiple of a and b is 

d x y, or m = d x y ; but dx = a, and y = -', substituting, we 

h 
have m = a X -; • 

d 

In a similar manner we could show that m — b X -;• 

d 

Hence the following 

RULE. 

Find the greatest common divisor of the two quantities ; di- 
vide one of the quantities by this, and multiply the quotient by 
the other quantity. 



LEAST COMMON MULTIPLE. 65 

Note. If there are three or more quantities, find the least common 
multiple of two of them, and then of that result and the third quantity ; 
and so on. 

1. Find the least common multiple of 6 x 2 — 17 x + 12 and 
12 a: 2 - 4 a -21. 

6 a; 2 - 17 x + 12)12 a: 2 - 4sc-21(2 
12 x 2 - 34 x + 24 



30 x - 45 
2 x - 3 ) 6 x 2 - 1 7 x + 1 2 ( 3 x - 4 
6 x 2 — 9 x 



- 8 x + 12 

- 8 a + 12 



Hence, 2 a: — 3 is the greatest common divisor of the two 
quantities ; dividing the first given quantity by this, we obtain, 
as a quotient, 3 x — 4 ; multiplying the second given quantity 
by this quotient, we have 

(3 a; -4) (12 a; 2 -4 a: -21), or 36 x 3 - 60 x 2 - 47 x + 84 
as the required least common multiple, Ans. 

EXAMPLES. 

Find the least common multiples of the following : 

2. 6 x 2 + 13 x - 28 and 12 x 2 - 31 x + 20. 

3. 8 x 2 + 30 x + 7 and 12 x 2 - 29 x - 8. 

4. a 3 + a 2_ 8 a _ 6 and 2 a 3 - 5 a 2 - 2 a + 2. 

5. 2 x 3 + x 2 - x + 3 and 2 .t 3 + 5 x 2 - x - 6. 

6. (t 3 -2« 2 H2ffii 2 - 6 3 and a 3 + a 2 6 - a b 2 - b 3 . 

7. x* + 2 x 3 + 2 x 2 + x and a « 3 — 2 a x — a. 

8. 2x 4 -llx- 3 +3a; 2 + 10a; and 3a; 4 - 14z 3 - 6ar+ 5oj. 



66 ALGEBRA. 



XL — FRACTIONS. 

133. A Fraction is an expression indicating a certain 
number of the equal parts into which a unit has been divided. 

The denominator of a fraction shows into how many parts 
the unit has been divided, and the numerator how many parts 
are taken. 

134. A fraction is expressed by writing the numerator 
above, and the denominator below, a horizontal line. Thus, 
- is a fraction, signifying that the unit has been divided into 

b equal parts, and that a parts are taken. 

The numerator and denominator are called the terms of a 
fraction. 

Every integer may be considered as a fraction whose denomi- 

a 
nator is unity ; thus, a = r- . 

135. An Entire Quantity is one which has no fractional 
part ; as, ab, or a — b. 

136. A Mixed Quantity is one having both entire and 

b a 

fractional parts ; as, a , or c + 



x + y 

137. If the numerator of a fraction be multiplied, or the 
de nominator divided, by any quantity, the fraction is multi- 
plied by that quantity. 

1. Let y denote any fraction ; multiplying its numerator by 

c, we have -— . Now, in - and — the unit is divided into b 
b b b 

equal parts, and a and a c parts, respectively, arc taken. Since 



FRACTIONS. 67 



c times as many parts are taken in — as in - , it follows that 

a c . .,. a 

— - is c times -. 

b o 

2. Let — denote any fraction ; dividing its denominator 

a a a 

by c, we have -. Now, in — and -, the same number of 

b be a b 

parts is taken ; but, since in — tbe unit is divided into 

a . hc 

b c equal parts, and in - into only b equal parts, it follows that 

• a . . , i , • a tt a 

each part m - is c times as large as each part m — . Hence, - 

is c times -r— . 
be 

138. If the numerator of a fraction be divided, or the de- 
nominator multiplied, by any quantity, the fraction is divided 
by that quantity. 

1. Let — denote any fraction ; dividing its numerator by c, 

we have-. Now, in Art. 137, 1, we showed that — was c 
a o a . ac . . b 

times -. Hence, - is — divided by c. 
b b . b 

2. Let - denote any fraction : multiplying its denominator 

b 

a a 

by c, we have — . Now, in Art. 137, 2, we showed that - was 

a bc a a . . h 

e times — . Hence, — is - divided by c. 
b c b c b 

139. If the terms of a fraction be both multiplied, or both 
divided by the same quantity, the value of the fraction is not 

altered. 

For, multiplying the numerator by any quantity, multiplies 
the fraction by that quantity ; and multiplying the denomi- 
nator by the same quantity, divides the fraction by that 
quantity. And, by Art. 44, Ax. 6, if any quantity be both 
multiplied and divided by the same quantity, its value is not 
altered. 



68 ALGEBRA. 

Similarly, we may show that if Loth terms are divided by 
the same quantity, the value of the fraction is not altered. 

140. We may now show the propriety of the use of the 
fractional form to indicate division, as explained in Art. 16 ; 

ft 

that is, we shall show that - represents the quotient of a di- 
vided by b. 

For, let x denote the quotient of a divided by b. 

Then, since the quotient, multiplied by the divisor, gives 
tbe dividend, we have b x = a. 

But, by Art. 137, bXj=a. 

Therefore, x = - . 

b 

141. A fraction is positive when its numerator and de- 
nominator have the same sign, and negative when they have 
different signs. 

For, a fraction represents the quotient of its numerator 
divided by its denominator ; consequently its proper sign can 
be determined as in division (Art. 91). 

142. The Sign of a fraction is the sign prefixed to its 
dividing line, and indicates whether the fraction is to be 
added or subtracted. 

Thus, in x -\ — — the sign + denotes that the fraction -j— , 

although essentially negative (Art. 91), is to be added to x. 

The sign written before the dividing line of a fraction is 
termed the apparent sign of the fraction ; and that de] tending 
upon the value of the fraction itself is termed the real sign. 

Thus, in -\ — — , the apparent sign is + , but the real sign 
is — . 

Where no signs are prefixed, plus is understood. 



ah 




— ah ah 




-ah 


h 




b -b 




-b 


ah 




a b — ah 




— ah 


— _ i 












b 




-b b 




-b 



FRACTIONS. 69 

143. From the principles of Arts. 140 and 141 we obtain, 

+ a; 

— a. 
— o 

From which it appears that, 

Of the three signs prefixed to the numerator, denominator, 
and dividing line of a fraction, any two may he changed with- 
out altering the value of the fraction ; hut if any one, or all 
three are changed, the value of the fraction is changed from 
+ to — , or from — to + . 

144. If either the numerator or denominator of the frac- 
tion is a polynomial, we mean by its sign the sign of the entire 
expression, as distinguished from the sign of any one of its 
individual terms • and care must be taken, pn changing signs 
in any such case, to change the sign before each term. 

„,, a — h—a + b b — a 

Thus ' -c^r-c^d> ov c—}v 

a—h—a+b h — a 
also, j = -., or . 

c — d —c + d d — c 

145. From Art. 141 we have 

abed _(-a)h (- c) (—d) __ a (- b) (- c) d 



efffh (~e)fgh e(-f)g(-h) 



, etc. ; 



abed _ (— a) bc{— d) _ a (— h) (— c) d 
~ Jff h - ( r e)fgY ~ e (-/) (- g) (- h)' : 

From which it appears that, 

If the terms of a fraction are composed of any number of 
factors, any even number of factors may have their signs 
changed without altering the value of the fraction • but if any 



70 ALGEBRA. 

odd number of factors have their signs changed, the value of 
the fraction is changed from + to — , or from — to +. 

a — b a — b b — a 



Thus, 



(x — y)(x — z)~ (y — x) (z — x)~ {y^- x) {x — z) 



b — a 1 n . b — a 

but does not equal 



(x — y)(z — x) (y — x){z — x) 



REDUCTION OF FRACTIONS. 

146. Reduction of Fractions is the process of changing 
their forms without altering their values. 

TO REDUCE A FRACTION TO ITS SIMPLEST FORM. 

147. A fraction is in its simplest form, when its terms are 
prime to each other. 

CASE I. 

148. When the numerator and denominator can be readily 
factored by inspection. 

Since dividing hoth numerator and denominator by the 

same quantity, or cancelling equal factors in each, does not 

alter the value of the fraction (Art. 139), we have the fol- 
lowing 

RULE. 

Resolve both numerator and denominator into their prime 
factors, and cancel all that are common to both. 

EXAMPLES. 

, ^, , 18 a 8 b 2 o . . , , 

1. ixeduce — — ^— 2 — ™ ^ s simplest form. 

18 a*b 2 c __ 2 . 3 . 3 . a . a . a . b . h . c 2ac 
45 a 2 b' 2 x ~ 5 . 3 . 3 . a . a . b . b . x 5 x ' * 

x 2 + 2 x — 15 

2. Eeduce — s — — to its simplest form. 

x- — 2x — 3 

g; 2 + 2x -15 _ (x + 5) (x - 3) _ ,-r + 5 
x 2 - 2 x - 3 : " (x + 1) (x - 3) ~ x~+l' 



FRACTIONS. 71 

_ _. , b c, — a r — b d + a d . . . 

3. Eeduce : ; — to its simplest form. 

a in — b in — an + n 

be — ac — bd + ad (b — a) (c — d) 

am — b m — an+ bn (a — b) (m — n) 

= (Art. 89) ("-f)(<*-«0 = fLll t Ans . 
(it — 0) (m — n) m — n 

Note. If all the factors of the numerator be removed by cancellation, 
the number 1 (being a factor of all algebraic expressions) remains to form a 
numerator. 

If all the factors of the denominator be removed, the result will be au 
entire quantity ; this being a case of exact division. 

Reduce the following to their simplest forms : 
4. SfUU. 13. 



c K.9U !IV IV - - 

oo mr n 6 

65x 2 y 3 z* 
2(> x* y° z~ 

„ 54 a 3 b 5 c 2 
72 a 2 b 2 c 

Wmxif ,« 

o. — — • 1 1 . 

to m x y- 

110 e 3 x 2 y 
9 - 22c 2 x 2 ■ 1S ' 

1A 2a 2 cd+2abcd 1Q 

1U - a~^ Tr — 7 ' iy> 

b <r x y + b ab,x y 

ii. 3»'-e« 4 y go. 

6x 2 y 2 — 12 xy 3 ac + ad — b c — b d 

19 x 2 — 2x — lh' 2mx + 3my — 2n 2 x — 3n 2 y 

x 2 + 10 x + 21 ' 2 m 2 x + 3 m 2 y—2nx—3n y 



m 2 — 10m 


+ 16 


m 2 + m - 


-72 * 


4 c 2 -20c 


' + 25 


25-4 


c 2 ' 


4 a — 


9 a n 2 



9bn 2 -12bn + 4:b 

8 x 3 + y 3 
4 x 2 — y 2 ' 

27 y 3 - 125 



25- 


-30 7/ + 9y 2 ' 


6 a; 2 


y — 2 x 3 y 


x 2 - 


-8x + 15 " 




4 — x 2 


X 3 - 


- 9 x 2 + 14 x ' 


a e 


— b c — ad + bd 



72 ALGEBRA. 

CASE II. 

149. When the numerator and denominator cannot be 
readily factored by inspection. 

Since the greatest common divisor of two quantities con- 
tains all the prime factors common to both, we have the fol- 
lowing 

KULE. 

Divide both numerator and denominator by their greatest 
common divisor. 

EXAMPLES. 

1. Reduce — - — 5 — — to its simplest form. 

6 cr — a — 12 

By the rule of Art. 126, we find the greatest common divisor 
of the numerator and denominator to he 2.a — 3. Dividing 
the numerator by this, the quotient is a — 1. Dividing the 
denominator, the quotient is 3 a + 4. Therefore, the simplest 

form of the fraction is -, Ans. 

3 a + 4' 

Reduce the following to their simplest forms : 

6a: 2 + a:-35 „ 6 a- 3 - 19 x 2 + 7 x + 12 



o -«-" " — "- — "J- g 

' '2u 2 -7a + 6' 

. 2 m 2 — 5 m + 3 Q 

' 12 m 3 - 28 m + 15 ' 

x a + x *-Sx-2 

O. -= ; s ~ 7Z • 1U. 



6 ' 2x 3 + 5x 2 -2:c + 3" 



8x 2 + 


22 a: + 5' 




10 a 2 - 


-a-21 




2u 2 - 


7 a + 6' 




2 m 2 


— 5 m + 


3 


12 m 2 - 


- 28 m + 


15 ' 


x s + x 2 — 3 x - 


-2 


x 3 — 4 


x 2 + 2x- 


4-3" 


6 a 3 — 


7 x 2 + 5 a 


•-2 



6a; 3 -25a: 2 + 17a- + 


20 


4a: 3 + 14ar+12a- + 


5 


4a; 3 -10a; 2 -12a'- 


rr' 

1 


12 a 2 + 16 a - 3 




10 a 2 + a - 21 ' 




x 3 — 4 x 2 + 4 a; — 1 




a 3 - 2 x 2 + 4 aj - 3 " 




6 x 3 — x 2 — 7 x — 2 





FRACTIONS. 73 

TO REDUCE A FRACTION TO AN ENTIRE OR MIXED QUANTITY. 

150. Since a fraction is an expression of division (Art. 
140), we have the following 

RULE. 

* 

Divide the numerator by the denominator, and the quotient 
will be the entire or mixed quantity required. 



EXAMPLES. 



ax — a 2 x 2 



1. Reduce — to an entire quantity. 

(ax — a 2 x 2 ) -i-ax = l — ax, Ans. 

q% A3 /j.3 

2. Reduce to a mixed quantity. 

b 3 

a — x)a 3 — x 3 — b 3 (a 2 + a x + x 2 , Ans. 

a — x 



a 3 — a 2 x 






a 2 x — 


x 3 — 


b 3 




9 

a x — 


a x 2 






ax 2 


— X 3 


— 


b z 


ax 2 


— x 3 







-b 3 

Reduce the following to entire or mixed quantities : 

ab — a 2 o~ 2 ■ k 

o. 

4. 

5. 

6. 

7. 

2ab 





b ' 




X 3 


+ y 3 




X 


+ y 




2: 


K 2 -3x-4: 






5 x 




X 3 


2 i rr 

— x' + 7x — 


6 




3 x 




a 2 


— 3ab + 4b 2 





8. 


x — 3 


9. 


X s — 1 
X-V 


10. 


4,x 2 -2x + 5 


2x 2 -x + \ ' 


11 


/Y»<* ___ ryi"— /y» i i _ */ 

*As *Aj *as *J 




X 2 + X — 1 


12 


2cc 3 -3.x 2 +4r-2 




2 x 2 — 3 x + 3 



74 ALGEBRA. 

TO REDUCE A MIXED QUANTITY TO A FRACTIONAL FORM. 

151. This is the converse of Art. 150; hence we may- 
proceed by the following 

RULE. 

Multiply the entire part by the denominator of the fraction ; 
add the numerator to the product when the sign of the fraction 
is + , and subtract it when the sign is — / writing the result 
over the denominator. 

EXAMPLES. 

a 2 £2 5 

1. Eecluce a + b — : — to a fractional form. 

a — b 

By the rule, 

a 2 _ b 2 _ 5 ( a + &) (g-b)- Q 2 - b°- - 5) 



a — b a — b 



a — b a — b 

Note. It will be found convenient to enclose the numerator in a pa- 
renthesis, when the sign before the fraction is — . 

Reduce the following to fractional forms: 

4 „ 3a 2 -2Z> 2 

2. x + l + .- 7.2a ^- 

x — 6 a a 

3. a + — 8. a 2 +ab + b 2 — 7 • 

n b — a 

4. 7a; - 4 " 2 + 5a - 9. 3z-2- 3 



~8 2x-l 

i x + 1 in / a * + b * 

5. * + i + __. 10. a -&_-^-. 

a +■ « a; — ^s 



FRACTIONS. 75 

TO REDUCE FRACTIONS TO A COMMON DENOMINATOR. 

t co 1 x> i 5cd 3mx j 3 n y 

152. 1. Keduce - — — , , and — -^ to a common 

3 crb 2 ab 2 ka 6 b 

denominator. 

Since multiplying each term of a fraction by the same quan- 
tity does not alter the value of the fraction (Art. 139), we 
may multiply each term of the first fraction by 4 a b, giving 

20 a b c d , « , 1 ■, -. „ . . 18 a 2 m x 

; each term of the second by b a , giving ; 

12 a? 6 2 ' J ' & ° 12« :i 6 2 

and each term of the third by 3 b, giving L. . 

12 a J b" 

It will be observed that the common denominator is the 
least common multiple of the given denominators, which is 
also called the least common denominator ; and that each term 
of either fraction is multiplied by a quantity which is obtained 
by dividing the least common denominator by its own denomi- 
nator. Kence the following 

RULE. 

Find the least common multiple of the given denominators. 
Divide this by each denominator, separately, and multiply the 
corresponding numerators by the quotients ; writing the results 
over the common denominator. 

Before applying the rule, each fraction should be in its sim- 
plest form ; entire and mixed quantities should be changed to 
a fractional form (Arts. 134 and 151). 

Note. The common denominator may be any common multiple of the 
given denominators. The product of all the denominators is evidently s 
common multiple, and the rule is sometimes given as follows : "Multiply 
'each numerator by all the denominators except its own, and write the 
results over the product of all the denominators." 

an a x x it 

2. Reduce - — — , — — , and . . ■ N , to a common de- 

1 — x (1 — x) 2 (1 — xy 

nominator. 



76 ALGEBRA. 

The least common multiple of the given denominators is 
(1 — be) 3 . Dividing this by the first denominator, the quotient 
is (1 — a:) 2 ; dividing it by the second denominator, the quo- 
tient is (1 — x) ; and dividing it by the third denominator, the 
quotient is 1. Multiplying the corresponding numerators by 
these quotients, we obtain a y (1 — x) 2 , a ar (1 — x), and x i/ 3 
as the new numerators. Hence the results are 

a y (1 — a?) 2 ax 2 (1 — x) x y 3 

(i-xy > {i-xy > and (i - xy ' Ans - 

EXAMPLES. 

Reduce the following fractions to a common denominator : 

a 3ab 2ac -,56c _ 4c — 1 3b — 2 1 5a 

3 - -^o— > -n— > and "To-- 6 - o „ , > K „ . , and 



8 ' 9 ' 12 ' 3ab ' 5ac ' 6b c 

. x 2 y xy z 7 y z 2 „ 2 3 4 

^** ~t , * i t~^ i and ^"tt — . /. — - — — , — -, and — -. 
10 ' 15 ' 30 a 3 a; 2 ' a a: 3 ' at 



a 2 x 



3y z Axz 5a;?/ 5 az 3bx ley — m 

b ' Yx-'Yy-' and TT " 8 - 6^' 87i' and 10* s 2 

9. -, — — -, and — 



a — b ' a + b ' a 2 + 6 2 ° 

10 # + 3 a? + 1 a; + 2 

a; 2 - 3 a: + 2' a; 2 - 5x + 6' x 2 - 4a; + 3' 

2a 3b 4c 

cr + a — 6 ' a 2 + 5a + 6' an< " a 2 — 4' 

12. T , — — T , and -j — T . 

a; — 1 ar— 1 x 3 — 1 

-n a & m — n a + b 

a m — b m + a n — b n' 2 a 2 — 2 a b ' 3 b m + 3 b n 

14. Reduce ; r^— - , — ;— - , and 



(a-b) (a-c) ' (b — a) (b-e)' (c-a)(c-b) 

to u common denominator. 



FRACTIONS. 77 

The fractions may be written (Art 145) as follows : 

, and 



(a - 6) (a - c) ' (a -b)(b- e) ' (a - c) (6 - c) 

The least common denominator is now (a — b) (a — e) 

(b — c). Applying the rule, we have the results, 

Q-a)(b-c) (h-l)(a-c) ^ ^ 



(a -b) (a- c) (b — c)' (a- b) (a - c) (b - c) 
(l-c)(a-b) 



(a -b){a- c) (b - c) 
Kecluce to a common denominator : 
,_ 3 2 .a — 2 

3 



, Ans. 



a 


— 


V 


a + 


V 




1 




2- 


X 


1 


+ 


X 

c 


> 
x — 

+ d 


1' 



16. z , 7, and 



x' 



1 — x . b — a 

17. -. j^-. pr-, -p. sT? ~ , and 



(a + b)(a-b)' (b-a)(c-d)' (d-c)(a + b) 

153. A fraction may he reduced to an equivalent one hav- 
ing a given denominator, by dividing the given denominator 
by the denominator of the fraction, and multiplying both terms 
by the quotient. 

1. Eeduce — ry to an equivalent fraction having 

ar — a b + b" 

a 3 + b 3 for its denominator. 

(a* + b 3 ) -j- (a 2 -ab + b 2 ) = a + b; 
multiplying both terms by a + b, 

a — b (a - b) (a + b) a 2 -b 2 

a*-ab + b* ~ (a 2 -ab + b 2 ) (a + b) ~ a 3 + b 3, **' 



78 algesra. 

EXAMPLES. 

2. Reduce to a fraction with a 2 — b 2 for its denom- 

a + b 

inator. 

x -\- 1 

3. Reduce - to a fraction with x 2 + 5 x — 24 for its 

x — 3 

denominator. 

. -r, , 3 m + 2 

4. Reduce to a fraction with G m? — 19 m + 10 

for its denominator. 

4 

5. Reduce to a fraction with a 3 — b 3 for its denom- 

a — 

inator. 

6. Reduce 1 + x to a fraction with 1 — x for its denomi- 
nator. 



ADDITION AND SUBTRACTION OF FRACTIONS. 

154. 1. Let it he required to add - to -. 

c c 

In -- and - , the unit is divided into c equal parts, and a 

and b parts, respectively, are taken, or in all a + b parts ; that 

a + 6 

is . Ihus, 

c 

a b a + b 

- + - =— -I— . 
c c c 

2. Let it he required to subtract - from - . 

c c 

The result must he such a quantity as when added to 7 will 

produce -; that quantity is evidently ■ • (Art. 154, 1). 

,„, a b a — b 

1 hus, = . 

c c c 



Hence the following 



FRACTIONS. 79 

RULE. 

To add or subtract fractions, reduce them, if necessary, to a, 
common denominator. Add or subtract the numerators, and 
write the result over the common denominator. 

The final result should be reduced to its simplest form, 
wherever such reduction is possible. 

3b - a b + a . 1 - 4 o 2 

1. Add — - , n 7 , and — - — — . 

3a ' 2b iab 

The least common multiple of the denominators is 12 a b. 
Then, by the rule of Art. 152, 

3b -a b + a 1 - 4 b* 12 b 2 - 4 ab 6 a b + 6 a 2 

+ "ITT" + , , = " -To-T- - + 



+ 



3a 2 b ±ab 12 a b 12 a b 

3 - 12 lr 12 fr 2 -4 a 6 + 6 a fr + 6 a 2 + 3-12 b 2 
12 a b 12 a b 

6a 2 +2ab + S 



12 a b 



, Ans. 



n n i 4 flj — 1 . 6 « — 2 

«. subtract — ^ from — . 

2 x 6 a 

The least common denominator is 6 ax. 

6a —2 4 a; — 1 12 a a; — 4 a: 12 ax — S a 



Then, 



3a 2a? Gaa; 6 a x 

12 a .r — 4 x — (12 ax — 3 a) 12 a a; — 4 a; — 12 a x + 3 a 

6 a x 6 ax 

3 a — 4 a; 



6 a a; 



, Ans. 



Note. When a fraction whose numerator is not a monomial is preceded 
by a - sign, it will be found convenient to enclose its numerator in a pa- 
renthesis before combining with the other numerators. If this is not done, 
care must be taken to change the sign of each term in the numerator before 
combining. 



80 ALGEBRA. 

4« 2 -l 3«i 2 -2 5« 2 c 2 +3 



3. Simplify 



2 a c 3 b 2 c 5 a 



3 



The least common denominator is 30 a b 2 c 3 . 

4a 2 -l _ 3 a b 2 - 2 _ 5a 2 c 2 +3 
2 a c 3 b 2 c 5 a c 3 

60 a 2 b 2 c 2 - 15 b' 2 c 2 _ 30 a 2 b 2 c 2 - 20 a c 2 30 a 2 b 2 c 2 + 18 If 
30 a b 2 c s 30 d b 2 c 3 30 a 6 a c 3 

_ 60 a 2 b 2 c 2 - 15 b 2 c 2 - (30 a 2 b 2 c 2 -20a c 2 ) - (30 a 2 b 2 c 2 +18 b ? ) 

30 a 6 2 c 3 

_ 60 a 2 & 2 c 2 - 15 6 2 c 2 - 30 a 2 b 2 c 2 + 20 a c 2 - 30 a 2 b 2 c 2 - 18 b 2 

30 a 6' 2 c 3 

20 a c 2 - 15 6 2 c 2 - 18 b 2 



30 a b 2 c 3 > AnS ' 



EXAMPLES. 

Simplify the following : 

. 2x — 5 3 a + 11 _ a — b 2a + b b — 3a 

4. \- . 9. 1 1 . 

12 18 4^6 8 

3 1 a 2 + 1 6 a 3 + 1 6-2 

* 5 a J 2 + 2a 2 i' '3 a 2 12 a 3 + ITT' 

2 a + 3 3 a + 5 2a;-l 2x + 3 6cc + l 

6 8 ' "12 ~H~ ^2(P" 

ra — 2 2 — 3?»,?i 2 m + 2 m + 2 m + 3 

' 2m » 3 m 2 n 3 ' ' ~J~ ~U~ ~2lT' 

b — 4a a + 5b 10 2 2x — 1 3.r 2 +l 

o. — — 1 — — - — . lo. 



24: a ^ 30b ' '3 6x 9 a; 2 ' 

,. a — 2 3cc + l 6 a: -5 3 

14. ■ -\ ' 

2 + 3 4 5 

3» + l 2&-1 4<?-l 6^+1 
12 a ~~8lT" + 16 c 24^' 



16. Simplify 



FRACTIONS. 81 

2x + l 3x — 1 11 



2 x (x — 1) 3 a; (a? + 1) 4 (a- 2 - 1) 
The least common denominator is 12 x (x 2 — 1). 
2x + l 3x — 1 11 



Then, 



2 x (x - 1) 3 x (x + 1) 4 (a; 2 - 1) 
6 (a + 1) (2 x + 1) 4 (as - 1) (3 x — 1) 33 a; 



12x(x 2 -l) 12x(x 2 -l) 12x(x 2 -l) 

12 x 2 + 18 x + 6 12 x- - 16 x + 4 33 x 



12 x (x 2 - 1) 12 x (x- - 1) 12 x (x 2 - 1) 

_ 12 x 2 + 18 x + 6 - (12 x 3 - 16 x + 4) - 33 x 
12 x (x 2 - 1) 

x + 2 



12 x (x 2 - 1) 



Aiis. 



Simplify the following : 

»;_* + *_. »i±» + s=» 

x + 2 3 — x a — 6 a + 6 

18. -i L_. 20.^-^. 

x + 7 x + 8 1 — xl + x 

a /> 2 a 5 

«1. — —7 H j H — o To • 

a + a — a' — b" 

1 1 2x 

22. + 



x + y x — // x' + y- 

1 x 3 

23. 5 -^ r + 



24. 



X — 1 x 2 — 1 X 3 — 1 

2 x — 6 x + 2 x + 1 



x 2 +3x + 2 x 2 -2x — 3 x 1 — x— 6' 



x x 2 x 



25. Simplify — — r + — h -5 



x + 1 1 — x x' 2 — 1 



82 ALGEBRA. 

The expression may be written (Art. 143) as follows : 

X X Li X 

+ 



X + 1 x — 1 X* — 1 
The least common denominator is ar — 1. 

ihen > ZTT^T-Z. T + ~2 7 = 31 i— Z72 T + Z2- 



x + 1 x — 1 ar — 1 a;' 2 — 1 x' —1 x 2 — 1 
or — a: — (.t 2 + cc) + 2 a; 



x 1 — 1 



Simplify the following : 



x* 



= 0, Ans. (Art. 102). 



3 4 

26. -2-+*. 28. -^-_+ — 



a — & b — a 3 x — a 2 # 2 — 9 

_,_. o rt -p x o a — J. -.-. x x x 

27. 1 — . 29. 1 

3a+3^2-2a 1+x 1-x 

1 1 1 

30. T-^f-Ti z + 



x 



31. 



(a — b){b — c) (b — a) (a — c) (c - a) (c — b) 
2 3 1 



(a; - 2) (a; - 3) (3 -*) (4 -a;) (a; - 4) (2 -a;) ' 



MULTIPLICATION OF FRACTIONS. 

155. We showed, in Art. 137, that a fraction could be 
multiplied by an integer either by multiplying its numerator 
or by dividing its denominator by that integer. We will now 
show how to multiply one fraction by another. 

Let it be required to multiply - by - . 

Let - = x, and - = y ; 

where x and y may be either integral or fractional. 



FRACTIONS. 83 

Since the dividend equals the product of the divisor and 

quotient. 

a = b x, and c = d y. 

Therefore, hy Art. 44, Ax. 3, a c = b d x y. 

Regarding a c as the dividend, b d as the divisor, and x y as 

the quotient, we have 

a c 

xy = Vd- 

Therefore, putting for x and y their values, 

a c a c 
~b X d^bd' 
Hence the following 

RULE. 

Multiply the numerators together for the numerator of the 
resulting fraction, and the denominators for its denominator. 

Mixed quantities should be reduced to a fractional form 
before applying the rule. 

When there are common factors in the numerators and 
denominators, they should be cancelled before performing the 
multiplication. 

EXAMPLES. 

- ,, 1A . -■ , 6x 2 y 10 a 2 y . 3b*x* 

1. Multiply together 5 — ^ , -=-= — - , and -. ~ . 

1 J & 5 a 3 b 2 ' 9 b x ' 4 a y 2 

6x 2 y 10 a 2 y 3 J 4 x s 6 x 10 X 3 a 2 b* x 5 y 2 b x* . 
5a 3 b 2 9bx A lay 2 ' 5 X 9 X 4 « 4 b 3 x y 2 ' ' a 2 ' " 

Multiply together the following : 

a 2 5 c -. a 3 b 2 . 3 abx 2 5 x y 2 



• — 2 and —3 — j • 4 - -5 — 2- ancl 5 — r - 

ra ?r m d n d bay 2, ab x 

3 a 3 x ,4fflJ _ m ?/ n , a x 

3 ' ^li- and KTZ, ■ 5 - 7^1 and 



7 A 4 5Am 4 « x m y n 



84 ALGEBEA. 

e 2a 6c ,5b 3ab 2 3ac 2 ,Sa<P 

o6 5a be 4cd 2&d 9&c 

„ 8 a-- 15 y 2 ,3 s; 4 _ 3 m* 2 n* . 11 z 2 

'• fT~3> i7 — 5 ? and in q • "• ,, o > o — .and-; — 5-. 

9 if lb z s ' 10 cc 3 2a; 2 '3??i' 4w 2 

10. Multiply together 

ar 2 _ 2cc x 2 -9 . a 2 + a; 

, and 



a;2_2a;-3' a; 2 -*' cc 2 +a:-.6' 

r-2x x 2 -9 cc 2 + ^ 

X-s X 



ce 2 — 2 cc — 3 x- — x x 2 + x — 6 

x(x-2)(x + 3) (a - 3) x (x + 1) x 

(x-3)(x + l)x(x- 1) (a + 3) (x -2) ~ x~^l ' 

Multiply together the following : 

1t 3.r 2 -cc , 10 

11. = and 



5 2cc 2 -4a;' 

,. 4 a; + 2 5cc 

1*. — ^ and 



2 x + 1 ' 

1Q a 2 -2ab + b 2 . b 

lo. ; and 



a + b ax — bx 

., . a — b . a 2 — b 2 

14. -=— -— and 



a 2 + a b a 2 — a b 

, _ 1 — x 2 1 — y 2 , 
10. q , s , and 



1 + 1/ ' X + X 2 ' 1 — X 

. x 2 -W . x 2 -25 
lb. „ — and -5 - — . 

X s + ox r-4a; 

a 3 — a 2 + a x s — 8 

17. -r and — 5 • 

x 2 + 2 x + 4 a s + 1 

,_ a: 2 + 5.r+G , x 2 — Ix 
18> x^I^^Yl and ^4' 



FRACTIONS. 85 



4 5 x 

19. 1 H — and — 



x x 1 x 2 — 8x + 7 



20. -4-1 and 



«X/ fcC 



2 -5cc + 6" 



a,a _ 3 x + 2 a 2 - 7 a; + 12 a 3 - 5 x 2 

2L aj»_8aj + 15' * 2 -5x + 4' and a* -4 ' 



22. ^ 2— j Lj ^r> and 1 + ^— 

a; — x y + y x~ + x y + y x — y 

a* _ U 1 - c 2 + 2 5 c a 2-p-c 2 -2bc 

i6 - a - + c--b 2 + 2ac a 2 + c 2 -b 2 -2ac' 

OA a + b a — b 4 b 2 a + b 

a — b a + b a — b Jo 

_„ 2x-\-y . y x 2 .. x 2 — y 2 

25. — — — 1 ^ = r- and -= — V- 

x + y y — cc x* — y* x' + y 



DIVISION OF FRACTIONS. 

156. We showed, in Art. 138, that a fraction could be 
divided by an integer either by dividing its numerator or by 
multiplying its denominator by that integer. We will now 
show how to divide one fraction by another. 

Let it be required to divide - by - . 

ft n 

Let x denote the quotient of --$-—. 

b a 

Then, since the quotient multiplied by the divisor gives the 
dividend, we have 

c a xe a 

XX d = b ] ° r ' ~d = b' 



86 ALGEBRA. 



Multiplying each of these equals by - (Art. 44, Ax. 3), 

G 



Therefore, 





a d 
x = 

be 


a 


c ad 
d be' 



Hence the following 



RULE. 

Invert the divisor, and proceed as in multiplication. 
Mixed quantities should he reduced to a fractional form, 



before applying the rule. 



EXAMPLES. 

6 aH , 9ab 3 



1. Divide r . , by - T7 r — = — r. 
5 x s y* J 10 x 2 y 5 



6a 2 b 9ab 5 6a 2 b 10 x 2 y h Any . 

: -^ y l_ — "L Ant 

5a: 3 ?/ 4 * 10.x 2 ?/ 5 5x s y** 9ab s 3b 2 x' 

x 2 — 9 x + 3 
2. Divide —zrs — by — p — . 

x 2 — 9 x + S _ (x + 3)(x- 3) 5 _ x — S 

~TE~ '' ~5~~~~ ~16~ ~ X x~+3~~ 3~ ' 

Divide the following: 

7 ??i 2 3n 2 x 2 — y 2 x 2 +xy 

6 - ~2~ y "13" ' ^-2^ + y 2 y ar-y * 

. 7« 3 i . 14 a b* _ n Sy 2 5 V 

1 J wr ?r o ra w x z — y x — y 



5. 


18 7??- a: 8 1 6 m 2 a: 4 
25 wy 2 } 5« 2 2/ 6 ' 


-1.1 

*■ a 2 +2a-15 } a 2 -2a-3' 


6. 


1 4t x 2 x 
4"^ by 12 + 3- 


in a; 3 — 4 a* , x 2 —3x-\-2 
' a; 2 + 5x + 6 b} « a +2aj-3' 



FRACTIONS. 87 



COMPLEX FRACTIONS. 

157. A Complex Fraction is one having a fraction in its 
numerator, or denominator, or both. It may be regarded as a 
case in division, since its numerator answers to the dividend, 
and its denominator to tbe divisor. 

However, since multiplying a fraction by any multiple of 
its denominator must cancel that denominator, to simplify a 
complex fraction, we may multiphj both of its terms by the 
least common multiple of their denominators. 

EXAMPLES. 

a 
1. Reduce - to its simplest form. 



FIRST METHOD. 

Proceeding as in division, 

a 

c a b a b 
-=^Xj=—t, Ans. 
a c a ca 

b 

SECOND METHOD. 

Multiplying both terms by the least common multiple of 
their denominators, 

a a 

-Xbc 
e c a o . 

-=- = —T) Ans. 

da, c a 

» b xbe 

a a 



2. Reduce — ^— — to its simplest form. 

b a 

a — b a + b 



88 ALGEBEA. 

The least common multiple of the denominators is a 2 — b 2 . 
Multiplying each term by a' 2 — lr, we have 

a (a -\-b) — a (a — b) cr + ab — a 2 + ab 2 a b 

Ans. 



b (a + b) + a (a — b) a b + b~ + a 2 — a b a 1 + b 2 ' 
3. Reduce - — to its simplest form. 

x 

1 1 X+ 1 33 + 1 

, Ans. 



1 1 _, a* a: + 1 + a; 2cc + 1 

.1 a; + 1 

1 +- 

Reduce the following to their simplest forms : 
4. -L-. 8. !-|. 12 



5. 


b 

a-\ — 
c 


m 

x 

n 




6. 


n 
»- g 



cc 



y-^ + 9 

7. -5=-4. 11. 15 





1 * 


1 + 






X 




1 


x 2 + 






X 




1 ' 


1 + 






X 


a 


b 


b 


a 


1 


1* 


b 


a 




1 


x 2 + 




2 


X- 


— 




2/ 



w + ?i 1 «- + ft b + .b 2 ' 



n 12 

x — 7 -\ 

9. j. 13. -, 

a_q 18 

a; + 3 

X 



10. 



14. 


1 

1 — X 


1 

1 + £C 


1 

=- -H 


1 

-= 



31 

1+ - c + 1 
3 — x 



SIMPLE EQUATIONS. 89 

a 2 + b 2 a 2 — lr in — n m 3 — n s 

, n a 2 — b 2 a 2 + b 2 10 m + n in 3 + n 3 

16. ; =— • !"• ~? '?• 

a + b a — b m + n rnr + nr 



+ 
a — b a + b m — n mr — n 



.2 



i7. x + y y , i9. 



•> «2 



x + 2 y x 2 " 

a; 



v/ a; + y a + x 

158. In Art. 42, we defined the reciprocal of a quantity 

as being 1 divided *by that quantity. Therefore the reciprocal 

in it 

of — = — = — ; or, the reciprocal of a fraction is the frac- 
n in m 

ii 
tion inverted. 



XII. — SIMPLE EQUATIONS. 

159. An Equation is an expression of equality between 
two quantities. Thus, 

x + 4 = 16 

is an equation, expressing the equality of the quantities x + 4 
and 16. 

160. The First Member of an equation is the quantity on 
the left of the sign of equality. The Second Member is the 
quantity on the right of that sign. Thus, in the equation 
x + 4 = 16, x + 4 is the first member, and 16 is the second 
member. 

The sides of an equation are its two members. 

161. An Identical Equation is one in which the two mem- 
bers are equal, whatever values are given to the letters in- 
volved, if the same value be given to' the same letter in every 
part of the equation ; as, 



90 ALGEBRA. 

2a + 2bc = 2(d + bc). 

162. Equations usually consist of known and unknown 
quantities. Unknown quantities are generally represented by 
the last letters of the alphabet, x, y, z; but any letter may 
stand for an unknown quantity. Known quantities are repre- 
sented by numbers, or by any except the last letters of the 
alphabet. 

163. A Numerical Equation is one in which all the known 
quantities are represented by numbers ; as, 

2 x — 11 = x — 5. 

A Literal Equation is one in which some or all the known 
quantities are expressed by letters j as, 

2x + a = bx 2 — 10. 

164. The Degree of an equation containing but one un- 
known quantity is denoted by the highest power of that 
unknown quantitj' in the equation. Thus, 

> are equations of the first degree. 
and c x = a' 2 + b d ' ) 

3 x 2 — 2 x = 65 is an equation of the second degree. 

In like manner we have equations of the third degree, fourth 
degree, and so on. 

When an equation contains more than one unknown quan- 
tity, its degree is determined by the greatest sum of the 
exponents of the unknown quantities in any term. Thus, 

x + x y = 25 is an equation of the second degree. 

a; 2 — y 2 z = a b 3 is an equation of the third degree. 

Note. These definitions of degree require that the equation shall not 
contain unknown quantities in the denominators of fractions, or under 
radical signs, or affected with fractional or negative exponents. 



SIMPLE EQUATIONS. 91 

165. A Simple Equation is an equation of the first degree. 

166. The Root of an equation containing' but one unknown 
quantity is the value of that unknown quantity; or it is tin- 
value which, being put in place of the unknown quantity, 
makes the equation identical. Thus, in the equation 

3x — 7 = x + 9, 
if 8 is put in place of x, the equation becomes 

24 - 7 = 8'+ 9, 
which is identical; hence the root of the equation is 8. 

Note. An equation may have more than one root. For example, in 

the equation 

x 2 = 7 J! -12, 

if 3 is put in place of x, the equation becomes 9 = 21-12; and if 4 is put 
in place of x, it becomes 16 = 28 — 12. Each of these results being iden- 
tical, it follows that either 3 or 4 is a root of the equation. 

167 It will be shown hereafter that a simple equation has 
but one root; an equation of the second degree, two mots; 
and, in general, that the degree of the equation and the num- 
ber of its roots correspond. 

168. The solution of an equation is the process of finding 
its roots. A root is verified, or the equation satisfied, when, 
the root being substituted for its symbol, the equation becomes 
identical. 

TRANSFORMATION OF EQUATIONS. 

169. To Transform an equation is to change its form with- 
out destroying the equality. 

170. The operations required in the transformation are 
based upon the general principle deduced directly from the 
axioms (Art. 44) : 



92 ALGEBRA. 

If the same operations are performed upon equal quantities, 
the results will be equal. 

Hence, 

Both members of an equation may be increased, diminished, 
multiplied, or divided by the same quantity, without destroy- 
ing the equality. 

TRANSPOSITION. 

171. To Transpose a term of an equation is to change 
it from one member to the other without destroying the 
equality. 

172. Consider the equation x — a = &. 
Adding a to each member (Art. 170), we have 

x — a + a = b + a 
or, x = b + a, 

where — a has been transposed to the second member by 
changing its sign. 

173. Again, consider the equation x + a = b. 
Subtracting a from each member (Art. 170), we have 

x + a — a = b — a 

or, x — b — a, 

where a has been transposed to the second member by chang- 
ing its sign. 

174. Hence the following 

EULE. 

Any term may he indisposed from one member of an equa- 
tion t<> the other, provided its sign be changed. 

Also, if the same term appear in both members of an equa- 
tion affected with tin same sign, it may be suppressed. 



SIMPLE EQUATIONS. 93 

1. In the equation 2x — 12 4- '3 = a; — 5a + 9, transpose 

the unknown terms to the first member, and the known terms 
to the second. 

Eesult, 2x — x + 5 a: = 12 — 3 + 9. 

EXAMPLES. 

Transpose the unknown terms to the first member, and the 
known terms to the second, in the following : 

2. 3x — 2a = 45+.2x. 

3. 4:X + 9 = 25-12x. 

4. 4 a 2 x + b 2 = — 4 a bx+±ac+ b 2 . 

5. a c + c x — a d = 2 a — 7 #. 

6. & c + a 2 x — m ?r = b x + a tZ — 5. 

7. 3 — & — x = c — 3x. 

8. 2a — 3c = 5ic — b — dx. 

9. 10 jc - 312 = 32 x + 21 - 52 x. 

CLEARING OF FRACTIONS. 

175. 1. Clear the equation -jr T = -X-+ s of fractions. 

o 4 o o 

The least common multiple of 3, 4, G, and 8 is 24. Multi- 
plying each term of the equation by 24 (Art. 170), we have 

16 x - 30 = 20 x + 9, 

where the denominators have been removed. Hence the fol- 
lowing 

RULE. 

Multiply each term of the equation by the least common 
multiple of the denominators. 



2, 


ax (I x m 


c — 




b en 




x 2 a 1 x 


:;. 






2 a 3 b lab 6 




ax ex a 


4. 


x i + , — o 




b a e 



94 ALGEBRA. 

Note. The operation of clearing of fractions may be performed by- 
multiplying each term of the equation by any common multiple of the de- 
nominators. The product of all the denominators is obviously a common 
multiple, and the rule is sometimes given as follows : "Multiply each term 
of the equation by the product of all the denominators." 

EXAMPLES. 

Clear the following equations of fractions: 

3x 5x b 

4 3 

7. x -%+ 20 = 1 + ^ + 26. 

n w X O ox n 

8 ' is-^-ir +! " l = () - 

—, *As its %K/ C\C\ f\ *-^ *^ *^ HO J-O </• 

b ' 5 + L2 = l0 _ "- " 12" 3T~ lo = 8~~6~- 

5 x 5 -\^ q x 9J i_j x 

10. Clear the equation 21 — ■ — == — — — - - t 

o 1G 2 

of fractions. 

The least common denominator is 16 ; multiplying each term 
by 1G, we have 

336 - (10 x - 10) = 11 - 3 x - (776 - 56 x) 
or, : !: U ; - lO x + 10 = 11 - 3 x - 776 + 56 x, Am. 

Note. When a fraction, whose numerator is not a monomial, is preceded 
by a —sign, it will he found convenient, on clearing of fractions, to enclose 
the numerator in a parenthesis. If this is not done, care must be taken 
to change the sign of each term in the numerator. 

Clear the following equations of fractions : 

11 x a + x 15 12 ax + b cx + d a 

2 3~~ ~~2' r be ' V 



SIMPLE EQUATIONS. 95 

13. -1 -1^ = 0. 15. ?_ ?___5a!_ 

1 + a; 1 — a? x + 1 a; — 1 ar — 1 

14 a ar — 3 1 _a 16 ,T ~*~ ^ a? — 3 2« + l_ 

2~2x + l _ 3 _ ' * ~5~ ~2~ ~3~ 



CHANGING SIGNS. 

176. The signs of all the terms of an equation may be 
changed without destroying the equality. 

For, in the equation a — x = b — c, let all the terms he 
multiplied by — 1 (Art. 170). Then, 

— a + x = — b + c 
or, x — a = c — b. 

For example, the equation — 5 x — a = 3 x — b, by chang- 
ing the signs of all the terms, may he written 

5x + a = b — 3x. 

SOLUTION OF SIMPLE EQUATIONS. 

177. To solve a simple equation containing hut one un- 
known quantity. 

1. Solve the equation 5 x — 7 = a? + 9. 

Transposing the unknown terms to the first member, and 
the known terms to the second, 

5 x — ^ = 7 + 9 
Uniting similar terms, 4 x = 16 

Dividing each member by 4 (Art. 170), 

x = 4, Ans. 

This value of x we may verify (Art. 168). Thus, substi- 
tuting 4 for x in the given equation, it becomes 

20 - 7 = 4 + 9, 

which is identical ; hence the value of x is verified. 



96 ALGEBRA. 

2. Solve the equation 8 x + 19 = 25 x — 32. 

Transposing, 8 x — 25 x = — 19 — 32 

Uniting terms, — 17 x = — 51 

Dividing by — 17, cc = 3, Ans. 

To verify the result, put 3 for x in the given equation. 

Then, 24 + 19 = 75 - 32 

or, 43 = 43. 

o a i .i .- 3x 5 2x x 

o. bolve the equation — — |- - = — . 

4 o *j 

Clearing of fractions, by multiplying each term of the equa- 
tion by 12, the least common multiple of the denominators, 

9x + 10 = 8x — Gx 

Transposing, 9a; — 8x + 6x = — 10 

Uniting terms, 7 x = — 10 

Dividing by 7, x = — — , Ans. 

To verify this result, put x = =r in the given equation. 

Then, _30 5_ _20 10 

_ .28 + 6~ - 21 + 14 






or, 



or, 



-90 + 70 _-80 + 60 
~84 _ "SlT 

_20_ _20 

84 " 84' 



RULE. 

Clear the equation of fractions if it has any. Transpose 
the unknown terms to tit e first member, and the known terms 
to the second, and reduce each member to its simples} firm. 
Diride both members of the resulting equation by the coefficient 
of the unknown quantity. 



SIMPLE EQUATIONS. 97 

EXAMPLES. 

Solve the following equations : 

4. 3sc + 5 = a; + ll. 7. 3x + 2-5ce = :b-7 + 3. 

5. 3z-2:=5x-lG. 8. 18-5cc-2x = 3 + a; + 7x. 

6. 2-2a; = 3-a?. 9. 5x^3 + 17 = 19-2x-2. 

10. Solve the equation 

5(7 + 3x)-(2a;-3)(l-2x)-(2a ; -3) 2 -(5 + : *0=O- 
Performing the operations indicated, we have 

35 + 15a: + 4a; 2 -8:r + 3-4x 2 + 12:r-9-5-a; = 

Transposing, and suppressing the terms 4 x 2 and — 4 x 2 , 

15 ;c-8;c + 12 a:-a; = - 35 -3 + 9 + 5 

18 x = - 24 

24 4 , 

x = -lS = -3> AnS - 

Solve the following equations : 

11. 3 + 2 (2x + S) = 2x -3(2 x + 1). 

12. 2cc — (4a;-l)=5a;-0-l). 

13. 7 («-2) -5 (a + 3) = 3 (2x- 5) -6 (4a;- 1). 

14. 3(3x + 5)-2(5z-3)=13-(5;c-16). 

15. (2 x - 1) (3 a; + 2) = (3 a; - 5) (2 a + 20). 

16. (5 - G a-) (2 x - 1) = (3 x + 3) (13 - 4 x). 

17. (^-3)' 2 -(5-a-) 2 = -4'a;. 

18. (2a;-l) 2 -3(^-2) + 5(3x-2)-(5-2a ; j 2 = 0. 

• 3 7 7 5 

19. Solve the equation „ — = zr^ — k~ -• 



98 ALGEBRA. 

Clearing of fractions, by multiplying each term by 12 x, the 
least common multiple of the denominators, 

36 - 42 = 7 x - 20 

- 7 x = - 36 + 42 - 20 

- 7 a; = - 14 

Solve the following equations : 

20. i^_7=— -— 24 ^— x -2x 8x 11 
4 3 4 ' * 5 *-^--2— -11. 

2i 1, 1 1 1 ok x -^ x _x 3x 

6 + 2^ _ 4 + 12^- Mt 2 + ~6~3 = 6~7T- 

«. |-| + | = 18. 26.*-f + 20 = | + | + 2 6. 

23. |_?_*=I_i 27. 2-^ = 7- 3 



345a? a; 2a; 2a;' 

oqqt n . • 3a; — 1 2 a; + 1 4a; — 5 

ao. holve the equation : = 4. 

4 3 5 

Multiplying each term by 60, 

45 x — 15 - (40 x + 20) - (48 x - 60) = 240 

45 x — 15 - 40 x — 20 — 48 x + 60 = 240 

45 x - 40 x — 48 x = 15 + 20 - 60 + 240 

- 43 a; = 215 

X = — 5, ^1?2S. 

Solve the following equations : 

oq q 5 ./■ + 3 7x OA 2 .r + 1 r 5 

29. 3>x-\ — = -s-. 30. x =— = 5x — -=. 

7-2 5 3 

31. 7 a- 7 = 3 x + 7. 



SIMPLE EQUATIONS. 99 



32. 2- 7 ^~^ = 3x 



33. 
34. 



6 4 

5a--2_3a- + 4 7x-\-2 _x — 10 
"IT ~T~ ~6~ ~2~ 

a; + 1 2 a; — 5 _ 11 x + 5 a; — 13 
~2~ "5" ~W ~3 ' 



5a- + l 17 x + 7 3.T-1 7 a; — 1 
4 + a- 3 a; — 2 11 a- + 2 2 - 9 a; 



36. 



14 



2 a- + 1 4 a- + 5 8 + x 2x + 5 
' ~3~~ ~T~ ~~6 _ ~8~ 

2 3 1 



38. Solve the equation 



x — 1 a 1 + 1 a'~ — 1 " 

Clearing of fractions, by multiplying each term by x 2 — 1, 
2 (a- + 1) - 3 (x - 1) = 1 
2a; + 2-3a- + 3 = l 

2a-3a; = -2-3 + l 
— a: = — 4 
x = 4, ^4«s. 
4 a- + 3 12 a- - 5 2 x - 1 



39. Solve the equation 



10 5 a; — 1 " 5 



Clearing of fractions partially, by multiplying each term 

by 10, 

120 a; -50 , 

4a; + 3 =- — t— = ±x — 2 

o .'' — 1 

_ , . 120 a; -50 

4a; + 3 — 4.r + 2 = — = -.— 

ox — 1 

: 120 x - 50 

5 = —= — 

o a- — 1 



100 ALGEBRA. 

Clearing of fractions, by multiplying each term by 5 x — 1, 

25 x - 5 = 120 x - 50 
25 a; - 120 x = 5 - 50 
- 95 £ = - 45 
_45_ 9 

x -y5-vJ> Ans ' 

Note. If the denominators are partly monomial, and partly polynomial, 
clear of fractions at first partially, multiplying by such a quantity as will 
remove the monomial denominators. 

Solve the following equations : 

1 —X 1 + x 1 — ar 
x — 1 x + 1 3 



X X 2 — 


5x 2 


3 j3 x - 


-7 3' 


2x-l 


2* + 7 


3a; + 4~ 


"3^ + 2 


5-2a 


3-2x 



41 — ' 4fj 

* ' x ._ 2 s + 2~x 2 -4' 

' ~x~+Y'' : ^+T* 9 = "3" " 1-dx ' 

„_ <6x 2 -3x + 2 _ ._ 2ar + 3a; 1 

43 - rr^ s s = 3. 47. -^ — — + — - = x + 1. 

2 a; 2 + 5 a; — 7 2 * + 1 3 <c 

48. Solve the equation 2 a ce — 3b = x + c — 3 ax. 
Transposing and uniting terms, 5 a x — x = 3b -\- c 
Factoring the first member, x (5 a — 1) = 3 b + c 
Dividing by 5 a — 1, a; = = r , ^4»s. 

49. Solve the equation (& — c .r) 2 — (a — c x)' 2 = b (b — a). 
Performing the operations indicated, 

b 2 -2bcx+ c 2 x 2 - a 2 + 2acx- c 2 x 2 = b 2 - a b 

Suppressing the term lr in both members, and the terms 
c 2 x 2 and — c 2 x 2 in the first member, 



SIMPLE EQUATIONS. 101 

— 2bcx — a 2 + 2acx = — ab 

2 a ex — 2b cx = a 2 — ab 

Factoring both members, 2 c x (a — b) = a (a — b) 

n . ,. a (a — b) a . 

Dividing by 2e(a-b), x= 2c ^_^ = ^ , Ans. 

Solve, the following equations : 

50. 2 ax + d — 3c — bx. 

51. 2 x — Aa = 3 ax + a 2 — a 2 x. 

52. 2 a x + 6 b 2 = 3 b x + 4 a b. 

53. 6 b m x — 5 a n — lo a ni — 2bnx. 

54. (or - 2 x) 2 = (4 x - b) (x + 4 e). 

55. (2 a - 3 x) (2 a + 3x) = b 2 -(3x- b) 2 . 

56. (3 a — x) (a + 2 x) = (5 a + x) (a — 2 x), 

3b x 2 _ 3 _ 2bx 

c a c 

- 3 +4^=3V 2 «( 2 - 3 «>- 



57. 




a 




X 


58. 


2a 


59. 


X 




2 


60. 


X 




a b 


61. 


X 




2 



1 + 2 ax 2x + \ 



2 a a 2 

X + " b X 

~~3lT = 3T~ 



(«-!)• 



I> c x x a c — 4 6 a; 



2b c 6 c 3b c 

62. Solve the equation .2 x - .01 - .03 x = .113 x + .161. 

FIRST METHOD. 

('banging the decimals into common fractions. 
2x 1 3x 118 a; 161 



10 100 100 " 1000 1000 



102 ALGEBRA. 

Multiplying each term by 1000, 

200 x - 10 - 30 x = 113 x + 161 
57 x = 171 
x = 3, Ans. 

SECOND METHOD. 

Transposing, .2 x — .03 x — .113 x = .01 + .101 
Uniting terms, .057 x = .171 

Dividing by .057, x = 3, ^4ms. 

Solve the following equations : 

63. .3x- .02 - .003 x = .7- .06 a- - .006. 

64. .001 x - .32 = .09 x - .2 x - .653. 

65. .3 (1.2 X -5)=U + .05 x. 

66. .7 (x + .13) = .03 (4x- .1) + .5. 

67. 3.3a;- \ =.la; + 9.9. 

.5 

2-3a 5a: _ 2x- 3 _ a- - 2 7 
"T5 — + L25~ ~9~ "378 ^ " 9 ' 

178. To prove that a simple equation ran have but one root. 

We have soon that every simple equation can be reduced to 
the form x = a. 

Suppose, if possible, that a simple equation can have two 
roots, and that ->\ and r., are the roots of the equation x = a. 
Then (Art. 168), 

r x — a, 

r 2 = a. 

Hence, t x = r 2 \ that is, the two supposed roots are identical. 
Therefore a simple equation can have but one root. 



PROBLEMS. 103 



XIII. — PROBLEMS 

LEADING TO SIMPLE EQUATIONS CONTAINING ONE 
UNKNOWN QUANTITY. 

179. A Problem is a question proposed for solution. 

180. The Solution of a problem by Algebra consists of 
two distinct parts : 

1. The Statement, or the process of expressing the condi- 
tions of the problem in algebraic language, by one or more 
equations. 

2. The Solution of the resulting equation or equations, or 
the process of determining from them the values of the un- 
known quantities. 

The statement of a problem often includes a consideration 
of ratio and proportion (Art. 21). 

181. Ratio is the relation, with respect to magnitude, 
which one quantity bears to another of the same kind, and is 
the result arising from the division of one quantity by the 
other. 

A Proportion is an equality of ratios. 
Thus, 

a : b, or - , indicates the ratio of a to b. 

a : b = c : d, is a proportion, indicating that the ratio of a 
to b, is equal to the ratio of c to d. 

In a proportion the relation of the terms is such that the 
product of the first and fourth is equal to the product of the 
second and third. 

ct c 
For, a : b = c : <% is the same as j — -, which, by clearing of 

fractions, gives ad = b c. 



1 04 ALGEBRA. 

182. For tlie statement of a problem no general rule can 
be given ; much must depend on the skill and ingenuity of the 
operator. We will give a few suggestions, however, which 
will be found useful : 

1. Express the unknoivn quantity, <>r one of the unknoivn 
quantities, by taw of the final letter* of the alphabet. 

1'. From tlie given conditions, find expressions for the other 
unknown quantities, if any, in the problem. 

3. Form on equation, by indicating the operations necessary 
to verify the values of -the unknown quantities, were they 
already known. 

4. Determine the value of the unknown quantity in the 
equation th US formed. 

Note. Problems which involve several unknown quantities may often 
be solved by representing one of them only by a single unknown letter. 

1. What number is that to which if four sevenths of itself 
be added, the sum w.ill equal twice the number, diminished by 
27? 

Let x = the number. 

4 x 
Then -=— = four sevenths of it, 

and 2x = twice it. 

4 x 

By the conditions, x -\ — — = 2 x — 27 

Solving this equation, x = 63, the number required. 

2. Divide 144 into two parts whose difference is 30. 

Let x = one part. 

Tli en. 144 — x = the other part. 

By the conditions, x — (144 — x) = 30 

Solving this equation, x = 87, one part. 

144 — #= ~>7, the other part. 



PROBLEMS. 105 

3. A is three times as old as B ; and eight years ago he was 
seven times as old as B. What are their ages at present ? 

Let x = B's age. 

Then, 3 x — A's age. 

Now, x — 8 = B's age, eight years ago, 

and 3 x — 8 = A's age, eight years ago. 
By the conditions, 3x — 8 = 7 (x — 8) 

Whence, x = 12, B's age, 

and, 3 x = 36, A's age. 

4. A can do a piece of work in 8 days, which B can perform 
in 10 days. In how many days can it he done hy both work- 
ing together ? 

Let x = the number of days required. 

Then, - = what both can do in one day. 

Also, — = what A can do in one day, 

o 

and j- = what B can do in one day. 

Since the sum of what each separately can do in one day is 
equal to what both can do together in one day, 

i JL -i 

8 + 10 _ x 

Whence, x = 4f , number of days required. 

5. A man has $ 3.64 in dimes, half-dimes, and cents. He 
has 7 times as many cents as half : dimes, and one fourth as 
many half-dimes as dimes. How many has he of each ? 



106 ALGEBRA. 

Let x = the number of dimes. 

x 

Then, - = the number of half-dimes, 

4 

7 x 
and — r— = the number of cents. 

4 

Now, 10 x = the value of the dimes in cents, 

and —j— == the value of the half-dimes in cents. 

4 

By the conditions, 10 x -\ ; — | -— = 364 

J '44 

Whence, x = 28, number of dimes, 

x 

- = 7, number of half-dimes, 

4 

— t— = 49, number of cents. 



6. Two pieces of cloth were purchased at the same price per 
yard ; but as they were of different lengths, the one cost $ 5 
and the other $ 6.50. If each had been 10 yards longer, their 
lengths would have been as 5 to 6. Required the length of 
each piece. 

Since the price of each per yard is the same, the lengths of 
the two pieces must be in the ratio of their prices, that is, as 5 
to 6h, or as 10 to 13. Therefore, 

Let 10 x = the length of the first piece in yards, 

and 13 x = the length of the second piece in yards. 

By the conditions, 10 x + 10 : 13 x + 10 = 5 : 6 

or (Art. 181), 6 (10 x + 10) = 5 (13 x + 10) 

"Whence, x = 2. 

Then, 10 x = 20, length of first piece, 

and 13 x = 26, length of second piece. 



PROBLEMS. 107 

7. The second digit of a number exceeds the first by 2 ; and 
if the number, increased by 6, be divided by the sum of the 
digits, the quotient is 5. Required the number. 

Let x = the first digit. 

Then, x + 2 = the second. 

Since the number is equal to 10 times the first digit, plus 
the second, 

10 x + x + 2, or 11 x + 2 = the number. 

11 x _|_ 2 + 6 

By the conditions, ■ ^— = 5 

J x + x + 2 

Whence, x = 2, the first digit, 

and x + 2 = 4, the second digit. 

Therefore the number is 24. 

8. Two persons, A and B, 63 miles apart, set out at the 
same time and travel towards each other. A travels 4 miles 
an hour, and B 3 miles. What distance will each have trav- 
elled when they meet ? 

Let x = the distance A travels. 

Then, 63 — x = the distance B travels. 

x 

- = the time A takes to travel x miles, 

and — — — = the time B takes to travel 63 — x miles. 
o 

By the conditions of the problem, these times are equal ; 
x 63 — x 

4=^r- 

Whence, x = 36, A's distance, 

and 63 — x = 27, B's distance. 



108 ALGEBRA. 

9. At what time between 3 and 4 o'clock are the hands of a 
watch opposite to each other ? 

Let M represent the position of 
the minute-hand at 3 o'clock, and H 
the position, of the hour-hand at the 
same time. 
\jj Let M 1 represent the position of 
Ih' the minute-hand when it is opposite 
to the hour-hand, and H 1 the po- 
sition of the hour-hand at the same 
time. 

Let x = the arc M H H' M', the space over which the min- 
ute-hand has moved since 3 o'clock. 

x 
Then, ^ = the arc H H', the space over which the hour- 

hand has moved since 3 o'clock. 

Also, the arc MH= 15 minute spaces, 

and the arc H' M 1 = 30 minute spaces. 

Now, arc M H H 1 M = arc MH+ arc H H< + arc H< M, 




x 
or, x = 15 + j^ + 30 

Solving this equation, x = 49 ^ minute spaces. 
That is, the time is 49-^ minutes after 3 o'clock. 

PROBLEMS. 

10. My horse and chaise are worth $ 336 ; but the horse is 
worth twice as much as the chaise. Required the value of 
each. 

11. What number is that from which if 7 be subtracted, one 
sixth of the remainder will be 5? 

12. What two numbers are those whose difference is 3, and 
the difference of whose squares is 51 ? 



PROBLEMS. 109 

13. Divide 20 into two such parts that 3 times one part may 
be equal to one third of the other. 

14. Divide 100 into two parts whose difference is 17. 

15. A is twice as old as B, and 10 years ago he was 3 times 
as old. What are their ages ? 

16. A is four times as old as B ; in thirty years he will be 
only twice as old as B. What are their ages ? 

17. A can do a piece of work in 3 days, and B can do the 
same in 5 days. In how many days can it he done by both 
working together ? 

18. A can do a piece of work in 3§ hours, which B can do 
in 2| hours, and C in 2i hours. In how many hours can it be 
done by all working together ? 

19. A and B can do a piece of work together in 7 days, 
which A alone can do in 10 days. In what time could B alone 
do it ? 

20. The first digit of a certain number exceeds the second 
by 4; and when the number is divided by the sum of the 
digits, the quotient is 7. What is the number '.' 

21. The second digit of a certain number exceeds the first 
by 3; and if the number, diminished by 9, be divided by the 
difference of the digits, the quotient is 9. What is the 
number ? 

22. A drover has a lot of oxen and cows, for which he gave 
$ 1428. For the oxen he gave $ 55 each, and for the cows $ 32 
each ; and he had twice as many cows as oxen. Required the 
number of each. 

23. A gentleman, at his decease, left an estate of $1872 for 
his wife, three sons, and two daughters. His wife was to re- 
ceive three times as much as either of her daughters, and each 
son to receive one half as much as each of the daughters. Re- 
quired the sum that each received. 



HO ALGEBRA. 

24. A laborer agreed to serve for 36 days on these condi- 
tions, that for every day he worked he was to receive $1.25, 
but for every day lie was absent he was to forfeit * 0.50. At 
the end of the time he received $ 17. It is required to find 

, how many days he labored, and how many days he was absent. 

25. A man, being asked the value of his horse and saddle, 
replied that his horse was worth $114 more than his saddle, 
and that g the value of the horse was 7 times the value of the 
saddle. What was the value of each ? 

26. In a garrison of 2744 men, there are 2 cavalry soldiers 
to 25 infantry, and half as many artillery as cavalry. Re- 
quired the number of each. 

27. The stones which pave a square court would just cover 
a rectangular area, whose length is 6 yards longer, and breadth 
4 yards shorter, than the side of the square. Find the area of 
the court. 

28. A person has travelled altogether 3036 miles, of which 
he has gone 7 miles by water to 4 on foot, and 5 by water to 
2 on horseback. How many miles did he travel in each 
manner ? 

29. A certain man added to his estate ^ its value, and then 
lost $ 760 ; but afterwards, having gained $ 600, his property 
then amounted to $ 2000. What was the value of his estate at 
first? 

30. A capitalist invested § of a certain sum of money in 
government bonds paying 5 per cent interest, and the re- 
mainder in bonds paying 6 per cent ; and found the interest 
of the whole per annum to be $180. Required the amount of 
each kind of bonds. 

31. A woman sells half an egg more than halt her eggs. 
Again she sells half an egg more than half her remaining 
eggs. A third time she does the same; and now she has sold 
all her eggs. How many had she at first ? 



PROBLEMS. HI 

32. What number is that, the treble of which, increased by 
12, shall as much exceed 54, as that treble is less than 144 ? 

33. A ashed B how much money he had. He replied, " If 
I had 5 times the sum I now possess, I could lend you $ 60, 
and then i of the remainder would be equal to h the dollars I 
now have." Required the sum B had. 

34. A, B, and C found a purse of money, and it was mutu- 
ally agreed that A should receive $ 15 less than one half, that 
B should have $13 more than one quarter; and that C should 
have the remainder, which was $ 27. How many dollars did 
the purse contain? 

35. A number consists of 6 digits, of which the last to the 
left hand is 1. If tins number is altered by removing the 1 
and putting it in the units' place, the new number is three 
times as great as the original one. Find the number. 

36. A prize of $ 1000 is to be divided between A and B, so 
that their shares may be in the ratio of 7 to 8. Required the 
share of each. 

37. A man has $ 4.04 in dollars, dimes, and cents. He has 
one fifth as many cents as dimes, and twice as many cents as 
dollars. How many has he of each ? 

38. I bought a picture at a certain price, and paid the same 
price for a frame ; if the frame had cost $ 1.00 less, and the 
picture $ 0.75 more, the price of the frame would have been 
only half that of the picture. Required the cost of the 
picture. 

39. A gentleman gave in charity $ 46 ; a part in equal por- 
tions to 5 men, and the rest in equal portions to 7 women. 
Now, a man and a woman had between them $8. What 
was given to the men, and what to the women ? 

40. Separate 41 into two such parts, that one divided by 
the other may give 1 as a quotient and 5 as a remainder. 



112 ALGEBRA. 

41. A vessel can be emptied by three taps ; by the first 
alone it could be emptied in 80 minutes, by the second in 200 
minutes, and by the third in 5 hours. In what time will it be 
emptied if all the taps be opened ? 

42. A general arranging his troops in the form of a solid 
square, finds he has 21 men over; but, attempting to add 
1 man to each side of the square, finds he wants 200 men to 
fill up the square. Required the number of men on a side at 
first, and the whole number of troops. 

43. At what time between 7 and 8 are the hands of a watch 
opposite to each other ? 

44. At what time between 2 and 3 are the hands of a watch 
opposite to each other? 

45. At what time between 5 and 6 are the hands of a watch 
together ? 

46. Divide 43 into two such parts that one of them shall be 
3 times as much above 20 as the other wants of 17. 

47. Gold is 19} times as heavy as water, and silver 10i 
times. A mixed mass weighs 4160 ounces, and displaces 250 
ounces of water. What proportions of gold and silver does it 
contain ? 

48. A gentleman let a certain sum of money for 3 years at 
5 per cent compound interest ; that is, at the end of each year 
there was added J,, to the sum due. At the end of the third 
year there was due him .$2315.25. Required the sum let. 

49. A merchant ha!s grain worth 9 shillings per bushel 3 and 
other grain worth 1.'! shillings per bushel, in what proportion 
must he mix 40 bushels, so that he may sell the mixture at 
10 shillings per bushel '.' 

50. A alone could perform a piece of work in L2 hours; A 
and C together could do it in 5 hours; and C's work is § of 
B's. Now. the work has to be completed by noon. A begins 
work at 5 o'clock in the morning; at what hour can he he 
relieved by B and ( '. ami the work- he just finished in time'.' 



SIMPLE EQUATIONS. H3 

51. A merchant possesses $5120, but at the beginning of 
each year he sets aside a fixed sum for family expenses. His 
business increases his capital employed therein annually at the 
rate of 25 per cent. At the end of four years he finds that his 
capita] is reduced to $3275. What are his annual expenses? 

52. At what times between 7 and 8 o'clock are the hands of 
a watch at right angles to each other ? 

53. At what time between 4 and 5 o'clock is the minute- 
hand of a watch exactly five minutes in advance of the hour- 
hand ? 

54. A person has 11^ hours at his disposal ; how far may 
he ride in a coach which travels 5 miles an hour, so as to re- 
turn home in time, walking back at the rate of oh miles an 
hour ? 

55. A fox is pursued by a greyhound, and is 60 of her own 
leaps before him. The fox makes 9 leaps while the greyhound 
makes but 6 ; but the latter in 3 leaps goes as far as the former 
in 7. How many leaps does each make before the greyhound 
catches the fox ? 

56. A clock has an hour-hand, a minute-hand, and a second- 
hand, all turning on the same centre. At 12 o'clock all the 
hands are together, and point at 12. How long will it be 
before the minute-hand will be between the other two hands, 
and equally distant from each ? 



XIV. — SIMPLE EQUATIONS 

CONTAINING TWO UNKNOWN QUANTITIES. 

183. If we have a simple equation containing two unknown 
quantities, as 3 x — 4 y = 2, we cannot determine definitely 
the values of x and y ; because, for every value which we give 
to one of the unknown quantities, we can find a corresponding 



114 ALGEBRA. 

value for the other, and thus find any number of pairs of values 
which will satisfy the given equation. 

Thus, if we put x = G, then 18 — 4 y = 2, or y = 4 ; 

t if we put x = — 2, then — 6 — 4 y = 2, or y = — 2 ; 
if we put a; = 1, then 3 — 4 ?/ = 2, or y = £ ; etc. 

And any of the pairs of values < " , I, ■! n i, ■< ~ , , 

etc., will satisfy the given equation. 

If we have another equation of the same kind, as 5x + 7?/=17, 
we can find any number of pairs of values which will satisfy 
this equation also. 

Now suppose we are required to determine a pair of values 
which will satisfy both equations. We shall find but one pair 
of values in this case. For, multiply the first equation by 5 ; 
thus, 

15 x -20 y = 10; 

and multiply the second equation by 3 ; thus, 

15 x + 21 y = 51. 

Subtracting the first of these equations from the second 
(Art. 44), we have 

41 y = 41, 

or, p = l. 

In the first given equation put y = l; then 3 x — 4 = 2, or 

3 x = 6 ; whence, x = 2. The pair of values \ ~-i\ satisfies 

both the given equations; and no other pair of values can be 
found which will satisfy both. 

184. Simultaneous Equations are such as are satisfied by 
the same values of their unknown quantities. 

185. Independent Equations are such as cannot be made 
to assume the same form. 



SIMPLE EQUATION'S. 115 

186. It is evident, from Art. 183, that two unknown 
quantities require for their determination two independent, 
simultaneous equations. When two such equations are given, 
it is our object to obtain from them a single equation contain- 
ing but one unknown quantity. The value of that unknown 
quantity may then be found; and by substituting it in either 
of the given equations we can find, as in Art. 183, the value of 
the other. 

ELIMINATION". 

187. Elimination is the process of combining simultaneous 
equations so as to obtain from them a single equation contain- 
ing but one unknown quantity. 

There are four principal methods of elimination : by Addi- 
tion or Subtraction, by Substitution, by Comparison, and by 
Undetermined Multipliers. 

CASE I. 

188. Ellin {nation by Addition or Subtraction. 

1. Given ox — 3 y = 19, and 7 x + 4 y — 2, to find the 
values of x and y. 

Multiplying the first equation by 4, 20 x — 12 y = 76 
Multiplying the second equation by 3, 21 x + 12 y = 6 

Adding these equations, 41 x = 82 

Whence, x = 2. 

Substituting this value in the first given equation, 

10-3y = 19 
-3y = 9 
y = -3. 

We might have solved the equations as follows : 

Multiplying the first by 7, 35 x - 21 y = 133 (1) 

Multiplying the second by 5, 35 x + 20 y = 10 (2) 

Subtracting (2) from (1), — 41 y = 123 

2/ = -3. 



116 ALGEBRA. 

Substituting this value of y in the first given equation, 

5 x + 9 = 19 
5 a- = 10 
x = 2. 

The first of these methods is elimination by addition ; the 
second, elimination by subtraction. 

RULE. 

Multiply the given equations, if necessary, by such numbers 
or quantities as will make the coefficient of one of the unknown 
quantities the same in the two resulting equations. Then, if 
the signs of the terms having the same coefficient arc alike, 
subtract one equation from the other ■ if unlike, add the two 
equations. 

This method of elimination is usually the best in practice. 

CASE II. 
189. Elimination by Substitution. 
Taking the same equations as before, 

5 x — 3 y = 19 (1) 

7x + 4t/= 2 (2) 

Transposing the term 7 x in (2), 4 y = 2 — 7 x 

2 7 x 

Dividing by 4, y = — _ (3) 

Substituting this value of y in (1), 

5*_3 (^=^) =19 

Performing the operations indicated, 



SIMPLE EQUATIONS. 117 

Clearing of fractions, 20 x — (6 — 21 x) = 76 
or, 20 x - 6 + 21 x = 76 

Transposing, and uniting terms, 41 x = 82 

Whence, x = 2. 

2 — 14 

Substituting this value in (3), y = — j — = — 3. 

• RULE. 

7v//r/ £/ie 7v//»e o/o/^e o/ £Ae unknown quantities in terms 
of the other, from cither of the given equations; and substi- 
tute this value for that quantity in the other equation. 

This method is advantageous when either of the unknown 
quantities has 1 for its coefficient. 

CASE III. 
190. Elimination by Comparison. 

Taking the same equations as before, 

5 x - 3 y = 19 (1) 

7x + ±y= 2 (2) 

Transposing the term — 3 y in (1), 5 x = 3 y + 19 

3//+ 19 ,,. 

or, a = — g (3) 

Transposing the term 4 y in (2), 7 a; = 2 — 4 y 

or, a: = — 

Placing these two values of x equal to each other (Art. 44), 

3// + 19 _ 2-4y 

5 7 

Clearing of fractions, 21 y + 133 = 10 — 20 y 



118 ALGEBRA. 

Transposing, and uniting terms, 41 y = — 123 
Whence, y — — 3. 

— 9 + 19 

Substituting this value in (3), x = p 

o 



RULE. 

Find the value of the same unknown quantity in terms of 
the other, from each of the given equations : ami form a new 
equation by placing these values equal to each other. 

CASE IV. 

191. Elimination by Undetermined Multipliers. 

An Undetermined Multiplier is a factor, at first undeter- 
mined, but to which a convenient value is assigned in the 
course of the operation. 

Taking the same equations as before, 

5x-3y = 19 (1) 

7 x + 4 y = 2 (2) 

Multiplying (1) by m, 5 m x — 3 m y = 19 m (3) 

Subtracting (3) from (2), 

7 x — 5mx + iy + 3m y — 2 — 19 m 
Factoring, x (7 — 5 m) + y (4 + 3 m) =2 — 19 m (4) 

Now, let the coefficient of y, 4 + 3 m = ; then 3 m = — 4, 

4 

or m = — Kj substituting this value of m in (4), 
o 

/_ 20\ „ 76 

n 7 +ir) = 2 +3 

Clearing of fractions, x (21 + 20) = 6 + 70 

41 x = 82 
x = 2. 



SIMPLE EQUATIONS. 119 

Substituting this value in (2), 14 + 4 y = 2 

4y = -12 
y = -3. 

We might liave let the coefficient of x in (4), 7 — 5m = 0; 

7 
then m would have been ■= ; substituting this value of m in (4), 

o 



y(±+ )=2-- 



Clearing of fractions, y (20 + 21) = 10 - 133 

41 y = - 123 
y = -3. 

Instead of subtracting (3) from (2), we migbt have added 
them and obtained the same results. Also, in the first place, 
we might have multiplied (2) by m, and either added the re- 
sult to, or subtracted it from, (1). 



RULE. 

Multiply one of the given equations by the undetermined 
quantity, m ; and add the result to, or subtract it from, the 
other given equation. 

In the resulting equation, factored with reference to the 
unknown quantifies, place the coefficient of one of the un- 
known quantities equal to zero, and find the value of ra. 
Substitute tins value of m In the equation, and the result trill 
be a simple equation containing but one unknown quantity. 

This method is advantageous in the solution of literal 
equations. 

2. Solve the equations. 

ax + b y — c (1) 

a f x + b'y = c' (2) 



120 ALGEBRA. 

Multiplying (1) by m, a m x + b m y = c m (3) 

Add (2) and (3), a' x + a m x + V y + b m y = d + cm 
Factoring, x (a' + a vi) + y (b 1 + b m) = d + c m (4) 

In (4), put the coefficient of y, V + b in, equal to zero. 

Then, b m — — V ; whence, m = — -. 

b 

Substituting this value of m in (4), 



a V \ , c b 



x \a' — ) = d 



Clearing of fractions, x (a' b — ab') =b d — b' e 

1, ( j _ y c 

Whence, x = — — . 

a' b — ab' 

In (4), put the coefficient of x, a' + a m, equal to zero. 
Then, a m = — a ' ; whence, m = ■ . 



a 



Substituting this value of m in (4), 



a'b\ a'e 

y\ h — '-) — c — 



a ' a 



Clearing of fractions, y (ab 1 — a 1 b) = ad — a' c 

a d — a' c 

Wh ence, y = —r l — . 

ab' — a' b 

Before applying either of the preceding methods of elimina- 
tion, the given equations should be reduced to their simplest 
forms. 

EXAMPLES. 

192. Solve, by whichever method may be most advanta- 
geous, the following equations : 

3. 3cc + 72/ = 33; 2x + ±y = 20. 

4. 7x + 2y = 31; 3 z ~ 4 ?/ = 23. 

5. 6x-3y = 27; 4;r-6y = -2. 



SIMPLE EQUATIONS. 121 

6. 7 x + 3y = -50; 2y-5x = U. 

7. 8y + 12 .£ = 116 ; 2x — y = 3. 

8. 11 x + 3 y = - 124 ; 2 x - 6 y = 56. 

9. 9a; + 4?/ = 22;27/ + 3cc = 14. 

10. ^+^f-8; -8x + 2y = -S0. 

11. 7z-2y = G;2a; + 22/ = -24. 

,«-.- ^ ,.„ 2^; 11?/ 5 

12. 11 y + G a; = 115 ; — — = — ■= . 

13. | a} + |y = ^ ; 10*-12y = -62. 

5 7 

14. _7a; + 47/ = -113; £ + -?, = -. 

15> 2~3-°' 4 + G- b - 

16. ^- y = 31; a; + ^ = 33. 

17. A^_^ = _30; aj + 7y = 119. 

< 3 a 

18. rc + 2// = .G; 1.7 x -y = . 58. 
3 ^ ?/ cc 2 ?/ 

19. -t r -?-? 1[g L=^.4y-8 aJ = lL 

2 T 

» + 3y 3 7 y - x 

' 2x-y 8' 2 + x + 2y ~ 

21. a x + b y = m ; c x + d y= n. 

22. m cc + ?i ?/ = r ; m' cc — n' y = /. 

no xy y x 

aw a c 



122 ALGEBRA. 



x v 1 x y 1 

** a + b^ a~b a 2 -b 2> a-b^ a + b a 2 -b 2 

9* X 19 Vm*. X S 2 V~ X -Q7 X + V 

25l -2~ 12 = 4 +8 ' 3~ 8 4T~~^ V 

2 2,3 2 

26. + = 1;— - = 0. 

03 + // X- — V/ 03 + 2/ ^ — i7 

2x + y _17 2y + 03 4 2 a? - y _ 2 y - a? 

4*. x- -g- - 12 4 "' 3 4 "-y " 3 ' 

2 03 3.7/ x + 2?/_ 5x — 6y 
28< "3 5 4~~ ~° 4~"' 

03 ox — y H x 



29. Solve the equations, 

6_3_ 
x y~ 

8 15 

- + — = -1 

« y 

Multiplying the first equation hy 5, 

30_15 = 2Q 

x y ' 

Adding this to the second given equation, 

" = 19 

03 

Clearing of fractions, 38 = 19 03 

Whence 03 = 2. 

Substituting this value in the first given equation, 

y 

Transposing, = 1 

Whence, y = — 3. 



SIMPLE EQUATIONS. 123 

Solve the following equations : 





3 15 2 3 


-1. 




30. 


x y 4 ' x y 




31. 


12 18 42 15 


8 


17 


x y 5 ' x 


y 


3 


32. 


11 7 3 4 8 
x y 2 ' x y 


-10. 





nn a b c d 

66. - + - = m ; -+- = ». 
x y x y 

fi 4 9 8 

34. — + f- = 4a5; -^ — = 3a 2 -4J a . 

<za; 6y 6a; ay 

oc m ?i n m ,o 

oO. 1 =zm + n; — | = iiv + n". 

nx my x y 



XV. — SIMPLE EQUATIONS 

CONTAINING MORE THAN TWO UNKNOWN QUANTITIES. 

193. If we have given three independent, simultaneous 

equations, containing three unknown quantities, we may coin- 
hine two of them hy the methods of elimination explained in 
the last chapter, so as to obtain an equation containing only 
two unknown quantities ; we may combine the third equation 
with either of the two former in the same way, so as to obtain 
another equation containing the same two unknown quantities. 
Then from these two equations containing two unknown quan- 
tities we may derive, as in the last chapter, the values of those 
unknown quantities. These values being substituted in either 
of the given equations, the value of the third unknown quan- 
tity may be determined from the resulting equation. 

The method of elimination by addition or subtraction is 
usually the most convenient. 



124 ALGEBRA. 

194. 1. Solve the equations, 

8x-dy-7z = -36 

12 x— y — 3z — 36 

6x-2y- z = 10 

Multiplying the first by 3, 21 x — 27 y - 21 z = — 108 (1) 
Multiplying the second by 2, 21 x — 2 y — 6 z = 72 (2) 
Multiplying the third by 4, 21 x — Sy — 4tz = 10 (3) 

Subtracting (1) from (2), 

or, 

Subtracting (3) from (2), 

or, 

Multiplying (5) by 3, 
Adding (1) and (6), 

Substituting this value in (5), 
Substituting the values of y and z in the third given equation, 

sc = 4. 

In the same manner, if we have given n feidependent, 
simultaneous equations, containing n unknown quantities, we 
may combine them so as to form a — 1 equations, containing 
n — 1 unknown quantities. These, again, may be combinnl 
so as to form n — 2 equations, containing n — 2 unknown 
quantities; and so on : the operation being continued until we 
finally obtain one equation containing one unknown quantity. 

RULE. 

Multiply the given equations, if necessary, by such numbers 
or quantities as trill make the coeffirirnt of one of the un- 
known quantities the same in the resulting equations. Cont- 
inue these equations by addition <>r subtraction, so as to form 



2oy + loz = ISO 




5y+ 3z= 36 


(4) 


6y- 2z= 32 




3y- z= 16 


(5) 


9y- 3z= 18 


(6) 


11//= 81 




y = 6. 




z = 2. 





SIMPLE EQUATIONS. 125 

a new set of equations, one less in number than before, and 
containing one less unknown quantity. Continue the opera- 
tion with these new equations ; and so on, until an equation is 
obtained containing nut- unknown quantity. 

Find the value of this unknown quantity. By substituting 
it in either of the equations containing only two unknown 
quantities, find the value of a second unknown quantity. By 
substituting these values in either of the equations containing 
three unknown quantities, find the value of a third unknown 
quantity j and so on, until the values of all arc found. 

Note. This rule corresponds only with the method of elimination by- 
addition or subtraction ; which, however, as we have observed before, is 
the best in practice. 

2. Solve the equations, 

u + x + y = 6 
u + x -\- z = 9 

u + y + z = 8 
x+y+z=7 

The solution may here be abridged by the artifice of assum- 
ing the sum of the four unknown .quantities to equal an auxil- 
iary quantity, s. Thus, 

Let u + x + y + z = s. 

Then we may write the four given equations as follows : 

s-z = 6 (1) 

s-y=9 (2) 

s-x = S (3) 

s-u=7 (4) 



Adding, 4 s — s = 30 

Whence, s = 10. 

Substituting the value of s in (1),,(~), (3), and (4), we obtain 

z ■= 4, y = 1, x = 2, and u = 3. 



126 ALGEBRA. 

EXAMPLES. 

Solve the following equations : 

3. x + y+z = 53; x + 2 y + 3s = 107; x + 3y + 4s = 137. 

4. 3x-y — 2z = -23\ G x + 2 y + 3 z = 15; 

4x + 3 y — z = — G. 

5. 5x — 3y+2z = 41i 2x + y-z = l7; 5x + ±y-2z = 3%. 

6. 7a; + 47/-2 = -50; 4x- 5 y-3 z = 20; 

x — 3y — 4:Z = 30. 

7. 3u + x + 2y — z = 22; 4x — y + 3 s = 35; 

4:u + 3x-2y=19; 2 u + 4y + 2 2 = 46. 

8. a; + ?/ = 2j a2 + s = 3; y + * = — 1. 

9. ?/ + s = a ; aj + £ = J ; a; + y = c . 

10. 4a;-4y = a + 4s; 6y — 2a; = a + 2s; 7s-y = fl + ^. 

11 2 + 3~4-~ 43 ' 3"I + 2-° 4 ' 4 + 2~3 == ~°°- 

12. 2a + 2 2 , + 3 = -17-2w; ?/ + 3*=-2; 4aj. + * = 13j 

^ + 3y = -14. 

13. a y -\- b x = c; c x + a z = b ; b z + c y = a. 



14 4 ?_ 6 81 _2_ _3_ 10 _7 

a: y 2~ = 2~' 3^" + 2l/~7 r i" Z "2 

8 6 4 .. 

+ — = 11. 



9 .'■ // 7 

.- 3 2 J_ J_ -1 A A -1 

°' 4x "37/" ; ' ""32/ + 2z~ ; 2z + 4:x~ 

16. x — ay+ cr z — a 8 ; x — by+b' 2 z = b :i ; x — c y + c' 2 z = c 



8 



PROBLEMS. 127 



17 y~ z x + % _ 1 x —y x — z —n. 

2 ~ 4 ~2' 5 6 



y+z x + y 
4 2 



-4. 



18. £^±1_ (2 _,) = 0; ^±-^ = 2a-cx 
a c 

(« + f) 2 -ac(2 + a; + s) = -y. 



XVI. — PROBLEMS 

LEADING TO SIMPLE EQUATIONS CONTAINING MORE 
THAN ONE UNKNOWN QUANTITY. 

195. In the solution of problems in which we represent 
more than one of the unknown quantities by letters, we must 
obtain, from the conditions of the problem, as many indepen- 
dent equations as there are unknown quantities. 

1. If 3 be added to both numerator and denominator of a 
certain fraction, its value is f ; and if 2 be subtracted from 
both numerator and denominator, its value is h- Required 
the fraction. 

Let 
and 

By the conditions, 





X- 


= the 


numerator, 




y-- 


= the denominator, 


X 


+ 3 


o 




V 


+ 3" 


o 
O 




X 


o 

w 


1 




y 


-2 


2 






x ■■ 


= ~> 


y = 12. 



Solving these equations, 

That is, the fraction is T v. 

2. The sum of the digits of a number of three figures is 13 ; 
if the number, decreased by 8, be divided by the sum of the 
second and third digits, the result, is 25; and if 99 be added 
to the number, the digits will be inverted. Find the number. 



128 ALGEBRA. 

Let x = the first digit, 

y = the second, 

and z — the third. 

Then, 100 x + 10 y + z = the nuniher, 
and 100 z + 10 y + x = the number with its digits inverted. 

By the conditions, x + y + z = 13 

100 x + 10 y + z-8 
— Zo 

100 x + 10 y + z + 99 = 100 z + 10 y + x 

Solving these equations, x = 2, the first digit, 

y = 8, the second, 



3, the third. 



That is, the number is 283. 



3. A crew can row 20 miles in 2 hours down stream, and 
12 miles in 3 hours against the stream. Required the rate 
per hour of the current, and the rate per hour of the crew in 
still water. 

Let x = rate per hour of the crew in still water, 

and y = rate jjer hour of the current. 

Then, x + y= rate per hour rowing down stream, 
and x — y = rate per hour rowing up stream. 

Since the distance divided by the rate gives the time, we 
have by the conditions, 

20 2 



x + y 
12 



= 3 



x—y 
Solving these equations, x = 7, and y = 3. 



PROBLEMS. 129 

PROBLEMS. 

4. A says to B, " If i of my age were added to § of yours, 
the sum would be 19^ years." " But," says B, " if § of. mine 
were subtracted from J of youx-s, the remainder would be 18£ 
years." Required their ages. 

5. If 1 be added to the numerator of a certain fraction, its 
value is 1 ; but if 1 be added to its denominator, its value is £. 
What is the fraction? 

6. A farmer has 89 oxen and cows ; but, having sold 4 oxen 
and 20 cows, found he then had 7 more oxen than cows. Re- 
quired the number of eacb at first. 

7. A says to B, " If 7 times my property were added to \ of 
yours, the sum would be % 990." B replied, " If 7 times my 
property were added to \ of yours, the sum would be $ 510." 
Required the property of each. 

8. If \ of A's age were subtracted from B's age, and 5 years 
added to the remainder, the sum would be 6 years ; and if 4 
years were added to \ of B's age, it would be equal to fe of A's 
age. Required their ages. 

9. Divide 50 into two such parts that % of the larger shall 
be equal to § of the smaller. 

10. A gentleman, at the time of his marriage, found that his 
wife's age was to his as 3 to 4 ; but, after they had been mar- 
ried 12 years, her age was to his as 5 to 6. Required their 
ages at the time of their marriage. 

11. A farmer hired a laborer for 10 days, and agreed to pay 
him $ 12 for every day he labored, and he was to forfeit % 8 
for every day he was absent. He received at the end of his 
time % 40. How many days did he labor, and how many days 
was he absent ? 

12. A gentleman bought a horse and chaise for % 208, and i 
of the cost of the chaise was equal to § the price of the horse. 
What was the price of each ? 



130 ALGEBRA. 

13. A and B engaged in trade, A with $ 240, and B with 
$96. A lost twice as much as B; and, upon settling their 
accounts, it appeared that A had three times as much remain- 
ing as B. How much did each lose ? 

14. Two men, A and B, agreed to dig a well in 10 days; 
hut, having labored together 4 days, 1! agreed to finish the 
job, which he did in 16 days. How long would it have taken 
A to dig the whole well ? 

15. A merchant has two kinds of grain, one at 60 cents per 
bushel, and the other at 90 cents per bushel, of which he 
wishes to make a mixture of 40 bushels that may be worth 
80 cents per bushel. How many bushels of each kind must 
he use '.' 

16. A farmer has a box filled with wheat and rye; seven 
times the bushels of wheat are 3 bushels more than four times 
the bushels of rye ; and the quantity of wheat is to -the quan- 
tity of rye as 3 to 5. Required the number of bushels of each. 

17. My income and assessed taxes together amount to $ 50. 
But if the income tax be increased 50 per cent, and the as- 
sessed tax diminished 25 per cent, the taxes will together 
amount to $ 52.50. Required the amount of each tax. 

18. A and B entered into partnership, and gained 8200. 
Now 6 times A's accumulated stock (capital and profit) was 
equal to 5 times B's original stock; and 6 times B's profit 
exceeded A's original stock by $200. Required the original 
stock of each. 

19. A boy at a fair spent his money for oranges. If he had 
got five more for his money, they would have averaged a half- 
cent less ; and if three less, a half-cent each more. J low many 
cents did he spend, and how many oranges did he get? 

20. A merchant has three kinds of sugar. He can sell 3 
lbs. of the first quality, 4 lbs. of the second, and 2 lbs. of the 
third, for 60 cents; or, he can sell 4 lbs. of the first quality, 
1 lb. of the second, and 5 lbs. of the third, for 59 cents ; or, he 



PROBLEMS. 131 

can sell 1 lb. of the first quality, 10 lbs. of the second, and 3 
lbs. of the third, for 90 cents. Required the price per lb. of 
each quality. 

21. A gentleman's two horses, with their harness, cost him 
$120. The value of the poorer horse, with the harness, was 
double that of the better horse; and the value of the better 
horse, with the harness, was triple that of the poorer horse. 
What was the value of each ? 

22. Find three numbers, so that the first with half the other 
two, the second with one third the other two, and the third 
with one fourth the other two, shall each be equal to 34. 

23. Find a- number of three places, of which the digits have 
equal differences in their order ; and, if the number be divided 
by half the sum of the digits, the quotient will be 41 ; and, if 
396 be added to the number, the digits will be inverted. 

24. There are four men, A, B, C, and D, the value of whose 
estates is $14,000; twice A's, three times B's, half of C's, and 
one fifth of J)'s, is $16,000; As, twice B's, twice C's, and two 
fifths of D's, is $18,000; and half of A's, with one third of 
B's, one fourth of C's, and one fifth of D's, is $ 4000. Re- 
quired the property of each. 

25. A and B are driving their turkeys to market. A says 
to B, " Give me 5 of your turkeys, and I shall have as many 
as you." B replies, " Give me 15 of yours, and then yours 
will be f of mine." How man}' had each ? 

26. A says to B and C, " Give me half of your money and I 
shall have $ 55." B replies, " If you two will give me one 
third of yours, I shall have $ 50." But C says to A and B, " If 
I had one fifth of your money I should have $50." Required 
the sum that each possessed. 

27. A gentleman left a sum of money to be divided among 
his four sons, so that the share of the eldest was \ of the sum 
of the shares of the other three, the share of the second J of 
the sum of the other three, and the share of the third I of the 



132 ALGEBRA. 

sum of the other three ; and it was found that the share of the 
eldest exceeded that of the youngest by $ 14. "What was the 
whole sum, and what was the share of each person? 

28. If I were to enlarge my field by making it .") rods longer 
and 1 rods wider, its area would be increased by 240 square 
rods; but if I were to make its length 4 rods less, and its 
width 5 rods less, its area would be diminished by 210 square 
rods. Required the present length, width, and area. 

29. A boatman can row down stream, a distance of 20 miles, 
and back again in 10 hours; and he finds that he can row 2 
miles against the current in the same time that he rows 3 miles 
with it. Required the time in going and in returning. 

30. A and B can perform a piece of work in days. A and 
C in 8 days, and B and C in 12 days. In how many "lays can 
each of them alone perform it ? 

31. A person possesses a capital of $30,000, on which he 
gains a certain rate of interest; but he owes $20,000, for 
which he pays interest at another rate. The interest which 
he receives is greater than that which he pays by $800. A 
second person has $35,000, on which he gains the second rate 
of interest ; but he owes $ 24,000, for which he pays the first 
rate of interest. The sum which he receives is greater than 
that which he pays by $ 310. What are the two rates of in- 
terest ? 

32. A man rows down a stream, which runs at the rate of 
o\ miles per hour, for a certain distance in 1 hour and 40 min- 
utes. In returning it takes him 6 hours and .'50 minutes to 
arrive at a point 2 miles short of his starting-place. Find the 
distance lie pulled down the stream, and the rate of his pulling. 

33. A train running from Boston to New York meets with 
an accident which causes its speed to be reduced to ,1 of what 
it was before, and it is in consequence 5 hours late. If the 
accident bad happened 60 miles nearer New York, the train 
would have been only one hour late. "What was the rate of 
the train before the accident ? 



PROBLEMS. 133 

34. A and B run a mile. A gives B a start of 44 yards 
and beats him by 51 seconds, and afterwards gives him a start 
of 1 minute 15 seconds and is beaten by 88 yards. In how 
many minutes can each run a mile ? 

35. A merchant has two casks, each containing a certain 
quantity of wine. In order to have an equal quantity in each, 
he pours out of the first cask into the second as much as the 
second contained at first ; then he pours from the second into 
the first as much as was left in the first; and then again from 
the first into the second as much as was left in the second, 
when there are found to be 1G gallons in each cask. How 
many gallons did each cask contain at first ? 

36. A and B arc building a fence 12G feet long; after three 
hours A leaves off, and B finishes the work in 14 hours. If 
seven hours had occurred before A left off, B would have fin- 
ished the work in 4§ hours. How many feet does each build 
in one hour ? 

GENERALIZATION OF PROBLEMS. 

196. A problem is said to be generalized when letters are 
used to represent its known quantities, as well as unknown. 

The unknown quantities thus found in terms of the known 
are general expressions, or formula', which may be used for 
the solution of any similar problem. 

197. The algebraic solution of a generalized problem dis- 
closes many interesting truths and useful practical rules, as 
may be seen from the consideration of the following : 

1. The sum of two numbers is a, and their difference is b ; 
what are the two numbers ? 

Let x = the greater number, 

and y = the less. 

By the conditions, x + y==a 

x — y = b 



134 ALGEBRA. 

, . , . a + b . , 

►solving these equations, x = — - — , the greater number, 

i a — b , , 

and y = — - — , the less. 

Hence, since a and b may have any value 'whatever, the 
values of x and y are general, and may be expressed as rules 
for the numerical calculations in any like case; thus. 

To find two numbers when their sum and difference are 
given, — Add the sum and difference, and divide by 2, for the 
greater of tin 1 two numbers; and subtract the difference from 
the sum, and divide by 2, for the less number. 

For example, if the sum of two numbers is 35, and their 
difference 13, 

the greater = — = 24, 

and the less = - — - — = 11. 

2. A can do a piece of work in a days, which it requires b 
days for B to perform. In how many days can it be done if A 
and B work together ? 

Let x = the number of days required. 

Then - = what both together can do in one day. 

Also, - = what A can do in one day, 

and y = what B can do in one day. 

By the conditions, - + - = - 
a b x 

Whence, x — - — , number of days required. 

Hence, to find the time for two agencies conjointly to ao- 



PROBLEMS. 135 

complish a certain result, when the times are given in which 
each separately can accomplish the same, — Divide the product 
of the given tunes by their sum. 

For example, if A can do a piece of work in 5 days, and B 
in 4 days, the time it will take them hoth working together 

•ill 5x4 20 o, , 
will be -z . = — = Z% days. 

5 + 4 9 J J 

3. Three men, A, B, and C, enter into partnership for a 
certain time. Of the capital stock, A furnishes m dollars; B, 
n dollars; and C, p dollars. They gain a dollars. "What is 
each man's share of the gain ? 

Let x = A's share. 

Then, since the shares arc proportional to the stocks, 



n ./• 



in 



= B's share, 



and == C's share. 



m 

ix x ty %c 

By the conditions, x -\ 1 — = a 

in in 

Whence, x = ■ , A's share. 

/// + n + p 

Then, = , B's share, 

m m + n + p * 

and — — = , C's share. 

m in + n + p 

Hence, to find each man's gain, when each man's stock and 
the whole gain are given, — Multiply the whole gain by each 
man's stock, and divide tlisproduct by the whole stock. 

For example, suppose A's stock $300, B's $500, and C's 
$800, and the whole gain $320. 



136 




ALGEBRA. 




Then, 


A's share 


320 x 300 96000 


= $60, 


"300 + 500 + 800" 1600 




B's share 


320 x 500 160000 


= $100, 




"300 + 500 + 800" 1600 




and C's share 


320 x 800 256000 


= $160. 



300 + 500 + 800 1600 

PROBLEMS. 

4. A cistern can he rilled by three pipes ; bj the first in a 
hours, by the second in b hours, and by the third in c hours. 
In what time can it he filled by all the pipes running together ? 

5. Using the result of the previous problem, suppose that 
the first pipe fills the cistern in 2 hours, the second in 5 hours, 
and the third in 10 hours. In what time can it be filled by 
all the pipes running together ? 

6. Divide the number a into two parts which shall have to 
each other the ratio of m to n. 

7. Using the result of the previous problem, divide the 
number 20 into two parts which shall have to each other the 
ratio of 3 to 2. 

8. A courier left this place n days ago, and goes a miles 
each day. He is pursued by another, starting to-day and 
going b miles daily. How many days will the second re- 
quire to overtake the first ? 

9. In the last example, if n = 3, a — 40, and b = 50, how 
many days will he required ? 

10. Required what principal, at interest at r per cent, will 
amount to the sum a, in t years ? 

11. Using the result of the previous problem, what principal, 
at 6 per cent interest, will amount to $3108 in 8 years? 

J.2. Required the number of years in which j> dollars, at r 
per cent interest, will amount to a dollars. 



DISCUSSION OF PROBLEMS. 137 

13. Using the result of the previous problem, in how many 
years will $262, at 7 per cent interest, amount to $472.91 ? 

14. A banker has two hinds of money. It takes a pieces of 
the first to make a dollar, and b pieces of the second to make 
the same sum. If be is offered a dollar for c pieces, how many 
of each kind must he give ? 

15. In the last example, if a = 10, b = 20, and c= 15, how 
many of each kind must he give ? 

16. A gentleman, distributing some money among beggars, 
found that in order to give them a cents each he should want 
b cents more; he therefore gave them c cents each, and had d 
cents left. Required the number of beggars. 

17. A mixture is made of a pounds of coffee at m cents a 
pound, h pounds at n cents, and c pounds at j> cents. Re- 
quired the cost per pound of the mixture. 

18. A, B, and C hire a pasture together for a dollars. A 
puts in m horses for t months, B puts in n horses for t' months, 
and C puts in^j horses for t" months. What part of the ex- 
pense should each pay ? 



XVII. — DISCUSSION OF PROBLEMS 

LEADING TO SIMPLE EQUATIONS. 

198. The Discussion of a problem, or of an equation, is the 
process of attributing any reasonable values and relations to 
the arbitrary quantities which enter the equation, and inter- 
preting the results. 

199. An Arbitrary Quantity is one to which any reason- 
able value may be given at pleasure. 

200. A Determinate Problem is one in which the given 
conditions furnish the means of finding the required quantities. 



138 ALGEBRA. 

A determinate problem leads to as many independent equa- 
tions as tin-re are required quantities (Art. 195). 

201. An Indeterminate Problem is one in which there are 
fewer imposed conditions than there are required quantities, 
and, consequently, an insufficient number of independent 

equations to determine definitely the values of the required 
quantities. 

202. An Impossible Problem is one in which the condi- 
tions are incompatible or contradictory, and consequently can- 
not be fulfilled. 

203. A determinate problem, leading to a simple equation 
involving only one unknown quantity, can be satisfied by but 
one value of that unknown quantity (Art. 178). 

An indeterminate problem, or one leading to a less number 
of independent equations than it has unknown quantities, may 
be satisfied by any number of values. 

For example, suppose a problem involving three unknown 
quantities leads to only two equations, which, on combining, 
give 

x — z = 10, 

or, x = 10 + z. 

Now, if we make z = 1, then x = 11 ; 
z — 2, then x = 12 ; 
z = 3, then x = 13. 

Thus, we may find sets of values without limit that will sat- 
isfy the equation. Hence, 

An indeterminate equation may have any manlier of so- 
lutions. 

204. When a problem leads to more independent equations 
than it lias unknown quantities, it is impossible. 

For, suppose we have a problem furnishing three indepen- 
dent equations, as, 

x = y+l 
y — 7 — x 
xy = 16 



DISCUSSION OF PROBLEMS. 139 

From the first two we find x = 4 and y = 3. But the third 
requires their product to be 10; hence the problem is im- 
possible. 

If, however, the third equation had not been independent, 
but derived from the other two, as, 

x y = 12, 

then the problem would have been possible; but the last equa- 
tion, not being required for the solution, would have been 
redundant. 



INTERPRETATION OF NEGATIVE RESULTS. 

205. In a Negative Result, or a result preceded by a — 
sign, the negative sign is regarded as a symbol of interpre- 
tation. 

Its significance when thus used it is now proposed to in- 
vestigate. 

1. Let it be required to find what number must be added to 
the number a that the sum may be b. 

Let x = the required number. 

Then, a + x = b 

Whence, x = b — a. 

Here, the value of x corresponds with any assigned values of 
a and b. Thus, for example, 

Let a = 12, and b = 25. 

Then x = 25 - 12 = 13, 

which satisfies the conditions of the problem ; for if 13 be 
added to 12,* or a, the sum will be 25, or b. 

But, suppose a = 30, and b = 24. 
Then, x = 24 - 30 = - 0, 



140 ALGEBRA. 

which indicates that, under the latter hypothesis, the problem 
is impossible in an arithmetical sense, though it is possible in 
the algebraic sense of the words "number," "added," and 
" sum." 

The negative result, — 6, points out, therefore, in the arith- 
metical sense, either an error or "// impossibility. 

But, taking the value of x with a contrary sign, we see that 
it will satisfy the enunciation of the problem, in an arithmeti- 
cal sense, Avhen modified so as to read : 

What number must be taken from 30, that the remainder 
may be 24 ? 

2. Let it be required to determine the epoch at which A"s 
age is twice as great as B's ; A's age at present being 35 years, 
and B's 20 years. 

Let us suppose the required epoch to be after the present 
date. 

Let x — the number of years after the present date. 

Then, 35 + x = 2 (20 + x) 

Whence, x = — 5, a negative result. 

On recurring to the problem, we find it so worded as to 
admit also of the supposition that the epoch is before the pres- 
ent date; and taking the value of x obtained, with the con- 
trary sign, we find it will satisfy that enunciation. 

Hence, a negative result here indicates that a wrong choice 
was made of two possible suppositions which the problem 
allowed. 

From the discussion of these problems we infer : 

1. That negatire results indicate cither an erroneous enun- 
ciation of a 'problem, or a wrong supposition respecting the 
quality of some quantity belonging to it. 

2. That we may form, when attainable, a 'possible problem 
analogous to that which involved the impassibility, or correct 



DISCUSSION OF PROBLEMS. 141 

the wrong supposition, by attributing to the unknown quan- 
tity in fin 1 equation a quality directly opposite to that 
which ltml lice// attributed to it. 

In general, it is not necessary to form a new equation, l»ut 
simply to change in the old one the sign of each quantity 
which is to have its quality changed. 

Interpret the negative results obtained, and modify the 
enunciation accordingly, in the following 



PROBLEMS. 

3. If the length of a field be 10 rods, and the breadth 8 rods, 
what quantity must be added to its breadth so that the con- 
tents may be 60 square rods ? 

4. If 1 be added to the numerator of a certain fraction, its 
value becomes | ; but if 1 be added to the denominator, it be- 
comes §. What is the fraction ? 

5. The sum of two numbers is 90, and their difference is 
120 ; what are the numbers ? 

6. A is 50 years old, and B 40 ; required the time when A 
will be twice as old as B- 

7. A and B were in partnership, and A had 3 times as much 
capital as B. When A had gained $ 2000. and B $ 750, A had 
twice as much capital as B. What was the capital of each at 
first ? 

8. A man worked 14 days, his son being with him 6 days. 
and received $39, besides the subsistence of himself and son 
while at work. At another time lie worked 10 days, and had 
his son with him 4 days, and received $28. What were the 
daily wages of each '.' 



142 ALGEBRA. 



XVIII. — ZERO AND INFINITY. 

206. A variable quantity, or simply a variable, is a quan- 
tity to which we may give, in the same discussion, any value 
within certain limits determined by the nature of the problem ; 
a constant is a quantity which remains unchanged throughout 
the same discussion. 

207. The limit of a variable quantity is a constant value 
to which it may be brought as near as we please, but which it 
can never reach. 

Thus, if 3 be halved, the quotient § again halved, and so on 
indefinitely, the limit to which the result may be brought as 
near as we please, but which it can never reach, is zero. And, 
in general, if any quantity be indefinitely diminished by di- 
vision, its limiting value is zero. 

208. If any quantity be indefinitely increased by multipli- 
cation or otherwise, its limiting value is called Infinity, and is 
denoted by the symbol co . 

209. It is evident, from the definition of Art. 207, that if 
two variable quantities are always equal, their limiting values 
will be equal. 

210. We will now show how to interpret certain forms 
which may be obtained in the course of mathematical opera- 
tions. 

. ' a a 

Let us consider the fraction - : and let - = a*. 

b b 



1. Interpretation of 

Let the numerator of remain constant, and the denomi- 

h 

nator be indefinitely diminished by division. By Art. l-'>7. it' 
the denominator is divided by any quantity, the value of the 



ZERO AND INFINITY. 143 

fraction is multiplied by that quantity ; hence the value of the 
fraction, x, increases indefinitely as b is diminished indefi- 
nitely. The limiting value of b being (Art. 207), the limit- 
ing value of - will be - ; and the limiting value of x is co 

b a 
(Art. 208). Now - and x being two variable quantities always 

equal, by xlrt. 209 their limiting values are equal ; or, 



a 



a 
2. Interpretation of — . 

00 



Let the numerator remain constant, and the denominator be 
indefinitely increased by multiplication. By Art. lo8, if the 
denominator is multiplied by any quantity, the value of the 
fraction is divided by that quantity ; hence x is diminished 
indefinitely by division as the denominator increases in- 
definitely. The limiting value of b being oo, the limiting 

ft ct 

value of - will be — : and the limiting value of x is 0. By 

b co 

Art. 209 these limiting values are equal ; or, 

GO 



Problem of the Couriers. 

211. The discussion of the following problem, commonly 
known as that of Clairaut, will serve to further illustrate 

the form -, besides furnishing us with an interpretation of the 

, 
form pr . 

Two couriers, A and B, are travelling along the same road, 
in the same direction, \V R, at the rates of m and n miles per 
hour respectively. If at any time, say 12 o'clock, A is at the 



144 ALGEBKA. 

point P, and B a miles from him at Q, when and where are 

they together '.' 

I ! 1 1 

R' P Q B 

Let #= the required time in hours-, 

and x = the distance A travels in the time t, or the dis- 
tance from P to the place of meeting. 
Then x — a = the distance B travels in the time t, or the dis- 
tance from Q to the place of meeting. 

Since the distance equals the rate multiplied by the time, 

x = m t 
x — a ==n 1 

Solving these equations with reference to t and x, 

a 



x = 



It is proposed now to discuss these values on different sup- 
positions. 

1. in > n. 

This hypothesis makes the denominator m — n positive; 
hence the values of both t and x are positive. That is, the 
couriers are together after 12 o'clock, and to the right of P. 

This interpretation corresponds with the supposition made. 
For, if A travels faster than B, he will eventually overtake 
him, and in advance of their positions at 12 o'clock. 

2. in < n. 

This hypothesis makes the denominator m — n negative; 
hence the values of both t and x are negative. Now. from 
what we have observed in regard to negative results < Art. 205), 
these values of t and x indicate that the couriers w< re together 
before 12 o'clock, and to the left of P. 



111 - 


- n 


m 


a 


m- 


- n 



ZERO AND INFINITY. 145 

This interpretation corresponds with the supposition made. 
For, if A travels more slowly than B, he will never overtake 
him ; but as they are travelling along the same road, they 
must have been together before 12 o'clock, and before they 
could have advanced as far as P. 

3. m = n. 
This hypothesis makes the denominator m — n equal to zero ; 

ft 77h CL 

so that the values of t and x become - and -j—, respectively; 

or, by Art. 210, t = oo and x — co . Since from its nature 
(Art. 208), go is a value which we can never reach, the values 
of t and x may be regarded as indicating that the problem is 
impossible under the assumed hypothesis. 

This interpretation corresponds with the supposition made. 
For, if the couriers were a miles apart at 12 o'clock, and were 
travelling at the same rate, they never had been and never 
would be together. 

Thus, infinite results Indicate the imjjossibility of a problem. 

4. a = 0, and m > n or m < n. 

By this hypothesis, the values of t and x each become 



m — n ' 



or (Art. 102), £ = and x = 0. That is, the couriers are to- 
gether at 12 o'clock, at the point P, and at no other time and 
place. 

This interpretation corresponds with the supposition made; 
for, if the distance between them at 12 o'clock is nothing, they 
are together at P ; but as their rates are unequal, they cannot 
be together after 12 o'clock, nor could they have been together 
before that time. 

5. a = 0, and m = n. 

By this hypothesis, the values of t and x each take the 

form^. 



146 ALGEBRA. 

Referring to the enunciation of the problem, we see that if 
the couriers were together at 12 o'clock, and were travelling 
at the same rate, they always had been, and always would be, 
together. There is, then, no single answer, or finite number 
of answers, to the problem in this case; and results of this 
form are therefore called indeterminate. 

Thus, a result - indicates indeterminathn. 

212. The symbol -, however, does not always represent an 
indeterminate quantity which may have any fin ite mine. Now, 
in the preceding problem the result - was obtained in conse- 
quence of two independent suppositions, one causing the nu- 
merator to become zero, and the other the denominator. We 
say independent^ because the quantity m — n can be equal to 
without necessarily causing the quantity a to become 0. And 

in all similar cases, we should find the result - susceptible of 
the same interpretation. 

But if the symbol - is obtained in consequence of the same 
supposition causing both numerator and denominator to be- 
come zero, it will be found to have a single definite limiting 
value. 

a 2 — b- 
Take, for example, the fraction — ; if b = a, this single 

supposition causes both numerator and denominator to become 
zero, and the fraction takes the form -. 

Now, dividing both terms by a — b, we have 

o?-V _ a + b 

a*-ab~ a ' { ' 

which equation is true so Ion*; as b is not equal to a. It is 
not necessarily true when b is equal to a, because the second 



ZERO AND INFINITY. 147 

member was obtained by dividing both terms of the first mem- 
ber by a — h (which divisor becomes when b = a), as we 
cannot speak of dividing a quantity by nothing. 

In (1), as b approaches a, the limiting value of the first 

member is --, and the limiting value of the second member 

is 2. Thus we have (Art. 209), Q = 2. 

Hence the limiting value of the fraction, as b approaches a, 
is 2. 

213. A proper understanding of the theory of indetermi- 

nation, and of the relation of zero to finite quantities, will lead 
to the detection of the fallacy in some apparently remarkable 
results. 

For example, let a = b 

Then a i = a b 

Subtracting //-, a 2 — J 2 = a b — b' 2 

Factoring, (a + b) (a — b)—b (a — b) (1) 

Dividing by a — b, a + b = b (2) 

But b = a; hence a + a = a 

then 2 a ■ = a 

or, 2=1 

The error was made in passing from (1) to (2). Equation 
(1) may be written 

a + b a — b 
b a — b 

Now, as b = a, the second member is an expression of the* form 
- . But we assumed in going from (1) to (2) that - — - = 1, 

" ' Cv " (J 

or that -- = 1 ; which we have seen in Arts. 211 and 212 is not 
necessarily the case, as it may have any value whatever. 



148 ALGEBRA. 



XIX. — INEQUALITIES. 

214. An Inequality is an expression indicating that one of 
two quantities is greater or less than the other ; as, 

a > b, and m < n. 

The quantity on the left of the sign is called the first mem- 
ber, and that on the right, the second member of the inequality. 

215. Two inequalities are said to subsist in the same sense 
when the first member is the greater or less in both. 

Thus, 

a > b, and c > d ; or 3 < 4, and 2 < 3, 

are inequalities which subsist in the same sense. 

216. Two inequalities are said to subsist in a contrary 
sense, when the first member is the greater in the one, and 
the second in the other. Thus, 

a > b, and c < d ; or x < y, and u > z, 

are inequalities which subsist in a contrary sense. 

217. In the discussion of inequalities, the terms greater and 
less must be taken as having an algebraic meaning. That is, 

Of" 11 !/ two quantities, a and b, a is the greater when a — b 
is positive, and a is the less when a — b is negative. 

Hence, a negative quantity must be considered as less than 
nothing; and, of two negative quantities, that is the greater 
which has the least number of units (Art. 49). Thus, 

> - 2, and - 2 > - 3. 

218. An inequality will continue in the same sense after 

the same quantity has been added to, or subtracted from, each 
member. 



INEQUALITIES. 149 

For, suppose a > b ; 

then, by Art. 217, a — b is positive ; consequently, 

(a + c) — (b + c) and (« — e) — (ft — c) 

are positive, since each equals a — b. Therefore, 

a + c > b + c, and a — c > S — c. 

Hence, it follows that a term may be transposed from one 
member of an inequality to the other, if its sign be changed. 

219. If the signs of all the terms of an inequality be 
changed, the sign of inequality must be reversed. 

For, to change all the signs, is equivalent to transposing 
each term of the first member to the second, and each term of 
the second member to the first. 

220. If two or more inequalities, subsisting in the same 
sense, be added, member to member, the resulting inequality 
will also subsist in the same sense. 

For, let 

a> b, a'> V, a"> b", 



then, by Art. 217, a — b, a' — b 1 , a" — b", are all positive ; 

and consequently their sum 

a + a' + a" + —b — b' — b"— 

or, (a + a' + a"+ ) - (b + V + b" + ) 

i> positive. Hence, 

a + a'+ a!'+ > b + b' + b" + 

221. If two inequalities, subsisting in the same sense, be 
subtracted, member from member, the resulting inequality will 
not always subsist in the same sense. 



150 ALGEBRA. 

For, let 

a > b, and a' > V ; 

* 

■ 

then a — b and a! — b 1 are positive ; but a — b— (a' — b'), or 
{a — a') — (b — &')> ma y l )e either positive, negative, or 0. 

That is, 

a — a'> b — b', a — a' < & — b', or a — a' = b — V. 

222. -'// inequality will continue in the same sense after 
each member has been multiplied or divided by the same posi- 
tive quantity. 

For, suppose a > 5 ; 

then, since « — J is positive, if m is positive, 

vi (a — b) and — (a — b) 

m K ' 

are positive. That is, m a— m b and are positive. 

in in 

Hence, 

7 i a b 
m a > ??i y, and — > — . 

in m 

223. If each member of an inequality he multiplied or di- 
vided by the same negative quantity, the sign of inequality 
must be reversed. 

For, since multiplying or dividing by a negative quantity 
must change the signs of all the terms, the sign of inequality 
must be reversed (Art. 219). 

224. The solution of an inequality consists in determining 
the limit in the value of its unknown quantity. 

This may be done by the application of the preceding prin- 
ciples. 

When, however, an inequality and an equation are given, 
containing two unknown quantities, the process of elimination 
will be required in the solution. 



INEQUALITIES. 151 

In verifying an inequality, if the symbols of the unknown 
quantities be taken equal to their respective limits, the ine- 
quality becomes an equation. 

EXAMPLES. 

225. 1. Find the limit of x in the inequality 

23 2x „ 
i x - -j > -g- + 5. 

Clearing of fractions, 21 a; — 23 > 2 x + 15 
Transposing, and uniting, 19 ic > 38 
Whence, x > 2, ^4«s. 

2. Find the limits of x in the inequalities, 

ax + 55x-5«i >a 2 (1) 

5x-7ftj; + 7«Ki' 2 (2) 

from (1), ax + 5b x > a 2 + 5 ab 

x (a + 5 b) > a (a + 5 b) 
x > a. 

From (2), bx — 7ax<b 2 — 7ab 

x (b - 7 a) < b (b - 7 a) 
x< b. 
Hence, x is greater than a, and less than b, Ant;. 

3. Find the limits of x and y in the following inequality and 
equation : 

4 x + 6 y > 52 (1) 

4 jc + 2 y = 32 (2) 

Subtracting (2) from (1), 4 ?/ > 20 

2/ > 5. (3) 

From (2), we have y = 16 — 2 x 



152 ALGEBRA. 

Substituting in (3), 16 - 2 x > 5 

-2x >-ll 
11 





-*>-- 2 




or (Art. 219), 


*<T 




Hence, 


y > 5, and # 





4. Given 5 a? — 6 > 19. Find the limit of cc. 

5. Given 2x-5 >25; 3x-7<2x+13. Find tin- 
limits of x. 

6. Given 3 z + 1 > 13 — x; 4cc — 7 < 2 a; + 3. Find an 
integral value of x. 

7. Given 5# + 3?/>46 — y; ?/ — z = — 4. Find the lim- 
its of x and y. 

ex c d x cl 

8. Given —- + dx — crf> T ; -5 c sc + c d < -^-. Find 

the limits of x. 

9. Given 2x + 3y<57; 2x + y = 32. Find the limits 
of x and y. 

10. A teacher being asked the number of his pupils, replied 
that twice their number diminished by 7 was greater than 29 ; 
and that three times their number diminished by 5 was less 
than twice their number increased by 16. Required the num- 
ber of his pupils. 

11. Three times a certain number, plus 16, is greater than 
twice that number, plus 24 ; and two fifths of the number, plus 
5, is less than 11. Required the number. 

12. A shepherd lias a number of slice]) such that three times 
the number, increased by 2, exceeds twice the number, in- 
creased by 61; and 5 times the number, diminished by 70, is 
less than 4 times the number, diminished by 9. How nianv 
sheep has he ? 



INVOLUTION. 153 



XX. - INVOLUTION. 

226. Involution is the process of raising a quantity to any 
required power. 

This may he effected, as is evident from the definition of a 
power (Art. 17), by taking the given quantity as a factor as 
many times as there are units in the exponent of the required 
power. 

227. If the quantity to he involved is positive, the signs of 
all its powers will evidently be positive ; but if the quantity is 
negative, all its even powers will be positive, and all its odd 
powers negative. Thus, 

(— a f =(-a)x (— a) X (— a) = + cr X (— a) = — a 3 , 

(- a y = (- a) X (- a) X (- a) x (- a) = (- a 3 ) X (- a) = + a\ 

and so on. 
Hence, 

Every even poiver is positive, and every odd power has the 
same sign as its root. 

INVOLUTION OF MONOMIALS. 

228. 1. Let it be required to raise 5 a" b c 3 to the fourth 
power. 

5 a 2 bc 3 x5a 2 bc s x5aHc 3 x5a 2 b c 3 = 625 a 8 b* c v2 , Ans. 

2. Raise —3m n 3 to the third power. 
(— 3 m n 3 ) X (— 3 m n 3 ) X (— 3 m n 3 ) = — 21 m 3 ?j 9 , Ans. 

RULE. 

Raise the numerical coefficient to the required power, and 
multiply the exponent of each letter by the exponent of the re- 
(j uired power ; making the sign of every wen poiver positive, 
and the sign of every odd power the same as that of its root. 



154 ALGEBRA. 

EXAMPLES. 

j 

Find the values of the following : 

3. (a 2 x) 2 . 7. (2x m )\ 11. (-2ab n xf. 

4. (-3 cr b) s . 8. (2ab 2 x 8 ) 5 . 12. (-7w*ra) 4 . 

5. (-ab 2 c 3 y. 9. (a 2 b 2 )\ 13. (5a 2 6 8 c 4 ) 8 . 

6. (a n b) m . 10. (-a 2 c 3 ) 3 . 14. (-6 a; 3 ?/ 7 ) 3 . 

A fraction is raised to any required power by raising both 
numerator and denominator to the required power. 
Thus, 

2x*\*_{ 2x*\ ( 2x 2 \ f 2x*\ 8 a; 6 

'3f) ~\ 3y) X \ 3y 3 l X \ 3y 3 )~~ 21 if 

Find the values of the following : 

15 . m\ it. L^r. ia i 2x ^ z 



b J * V 3b I ' V 36 

16. R**\\ 18. fL'^V. 20. f-*'^' 



4 x y*/ \o / V 4 a~ 



INVOLUTION OF POLYNOMIALS. 

229. Polynomials may he raised to any power, as is obvious 
from Art. 226, by the process of successive multiplications. 

Thus, 

(a + b) 2 =(a + b)(a + b) = a 2 + 2ab + b' 2 , 

(a + b) 3 =(a + b) (a + b) (a + b) = a 3 + 3 a 2 b + 3 a b 2 + b 3 , 
and so on. Hence the following 

RULE. 

Multiply the polynomial by itself, until it has been taken as 
a factor as many times as there are units in the exponent of 
the required power. 



INVOLUTION. 155 

EXAMPLES. 

Find the values of the following : 
1, (a -b) s . 3. {l + a 2 + b 2 )\ 5. (a m -a n )\ 



2. (|-^) 2 - 4. (a + m-nf. 6. (a + b) 



5 



In Chapter XXXVII will be given a method for raising a 
binomial to any required power, without going through with 
the process of actual multiplication. 

SQUARE OF A POLYNOMIAL. 

230. It has been shown (Arts. 104 and 105) that the 
square of any binomial expression can be written down, with- 
out recourse to formal multiplication, by application of the 
formulae 

(a + b) 2 = a 2 + 2ab + b 2 , 

(a-by = a?-2ab + b 2 . 
We may also show, by actual multiplication, that 
(a + b + c) 2 =a 2 + 2ab + 2ac + b 2 + 2bc + c 2 , 

(a + b + c + d) 2 = a 2 + 2 ab + 2 ac + 2 ad + b 2 + 2 bc+ 2bd 

+ c 2 + 2cd + d 2 , 
and so on. 

These residts, for convenience of enunciation, may be writ- 
ten in another form, 

(a + b) 2 = a 2 + b 2 + 2ab, 

(a — b) 2 = a 2 + b 2 — 2 a b, 

(a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2 a c + 2b c, 

(a + b + c + d) 2 = a 2 + b 2 +c 2 +d 2 +2ab + 2ac + 2ad 

+ 2bc + 2bd + 2cd, 

and so on. Hence, the following 



156 ALGEBRA. 

RULE. 

Write the square of each term, together with twice its prod- 
uct by each of the terms following it. 

1. Square x 2 — 2 x — 3. 

Square of each term, a? 4 + 4 x 2 +9 

Twice x 2 X the terms following, — 4. x 3 — 6x 2 

Twice — 2 x X the term following, + 12 x 

» 

Adding, the result is x A — 4:X 3 — 2x 2 +12x + 9. 



EXAMPLES. 

Square the following expressions : 

2. a — b + c. 8. 1 + x + x 2 + x 3 . 

3. 2x 2 + 3x + 4. 9. x 3 -4x 2 -2x-3. 

4. 2x 2 - 3x + i 10. 2x 3 +x 2 +l x-l. 

5. a — b — c + d. l\. x 3 + bx 2 — x + 2. 

6. »i 3 + 2a; 2 + a;+2. 12. 3x 3 -2 x 2 -x+ ±. 

7. 1 — 2 a; + 3 ar. 13. a + & — c — d + e. 

CUBE OF A BINOMIAL. 
231. Hy actual multiplication we may show, 

(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 , 
(a-b) 3 = a 3 -3a 2 b + 3a b 2 - b 3 . 
Hence, for finding the cube of a binomial, the following 

RULE. 

Write the cube of the first term, phis three times the square 
of the first term times the second, 'plus three times the first 
term times the square of the second, plus the cube of the second 
term. 



3\3 



INVOLUTION. 157 



EXAMPLES. 

1. Find the cube of 2 x 2 - 3 y 3 . 
(2 a- 2 ) 3 + 3 (2 x 2 ) 2 (- 3 y 3 ) + 3 (2 a: 2 ) (- 3 y 3 ) 2 + (- 3 if) 
' = 8z 6 + 3 (4* 4 ) (-3t/ 3 ) + 3 (2 a; 2 ) (9 if) + (-27 y 9 ) 

_ 8 x G - 36 x* f + 54 x 2 if - 27 y 9 , Jns. 

Cube the following : 

2. a 2 +2b. 4. 3 a; -4. 6. 4 a; 2 -ay. 

3. 2m + 5«. 5. 2z 3 -3. 7. 3a;7/ + 5a6 



.2 



CUBE OF A POLYNOMIAL. 
232. By actual multiplication we may show, 

(a + ft + e y = a 3 + b 3 + c 3 + 3 a 2 b + 3 a 2 c + 3 b 2 a + 3 b 2 c 
+ 3 c 2 a + 3c 2 + Gabc, 
( a + l ) + c + d) 3 = a 3 + b 3 + c 3 + d 3 + 3a 2 b + 3a 2 c + 3a 2 d 

+ 3b 2 a + 3 b 2 c + 3b 2 d + 3 c 2 a + 3 c 2 6 
+ 3 c 2 d + 3 d 2 a + 3d 2 b + 3d 2 c+6abc 
+ 6 a b d + 6 a c d + 6 b c d, 
and so on. Hence, for finding the cube of a polynomial, the 
following 

BULE. 

Write the cube of each term, tor/ether with three times the 
product of its square by each of the other terms, and also six 
times the product of every three different terms. 

EXAMPLES. 

1. Find the cube of 2 a; 2 .— 3 x — 1. 

8 x 6 -21 x 3 - 1 

- 36 x h - 12 a 4 

-f 54 a; 4 - 27 x 2 

+ Qx 2 -9x 
+ 36x 3 



8 x 6 - 36 x s + 42 x 4 + 9 a' 3 - 21 x 2 - 9 x - 1, Arts. 



158 ALGEBRA. 

Find the cubes of the following : 

2. a + b — c. 5. 2 - 2 .r + x\ 

3. a; 2 -2-1. 6. 1 + a; + x 2 + X 3 . 

4. a- 6 + 1. 7. 2.« 3 -r + 2x-3. 



XXI. — EVOLUTION. 

233. Evolution is the process of extracting any required 

root of a quantity. 

This may be effected, as is evident from the definition of a 
root (Art. 17), by determining a quantity which, when raised 
to the proposed power, will produce the given quantity. It is, 
therefore, the reverse of involution. 

234. Any quantity whose root can be extracted is called a 
perfect power ; and any quantity whose root cannot be ex- 
tracted is called an imperfect power. 

A quantity may be a perfect power of one degree, and not of 
another. Thus, 8 is a perfect cube, but not a perfect square. 

235. To extract any root of a simple quantity, the expo- 
nent of that quantity must be divided by the index of the root. 

For, since the ?ith power of a m is a mn (Art. 228), it follows 
that the nth. root of a mn is a m . 

236. Any root of the product of two or more factors is 
equal to the product of the same root of each of the factors. 

For, we have seen in Art. 228, in raising a quantity com- 
posed of factors to any required power, that cadi of the factors 
is raised to the same power. 

237. From the relation of a root to its corresponding 
power, it follows, from Art. 227, that 



EVOLUTION. 159 

1. The odd roots of any quantity have the same sign as the 
quantity. 

Thus, \j a 3 = a ; and ^ — a 5 = — a. 

2. The even roots of a 'positive quantity are either positive 
or negative. 

For either a positive or negative quantity raised to an even 
power is positive. Thus, 

y/ a 4 = a or — a ; or, y' a* = ± a. 

Note. The sign ±, called the double sign, is prefixed to a quantity 
■when we wish to indicate that it is either + or - . 

3. Even roots of a negative quantity are not possible. 

For no quantity raised to an even power can produce a neg- 
ative result. Such roots are called impossible or imaginary 
quantities. 

EVOLUTION OF MONOMIALS. 

238. From the principles contained in Arts. 235 to 237, 
we obtain the following 

RULE. 

Extract the required root of the numerical coefficient, ami 
divide the exponent of each letter by the index of the root; 
making the sign of every even root of a positive quantity ±, 
and the sign of every odd root of any quantity the same as 
that of the quantity. 

If the given quantity is a fraction, it follows from Art. 228 
that we may fake the required root of both of its terms. 

EXAMPLES. 

1. Find the square root of 9 a 4 b- c 6 . 



\/9a 4 i 2 c 6 = ±3fl 2 i c s , Ans. 



2. Find the cube root of - 64 a 9 x z y 6 . 

.3/ 



^ _ 64 a 9 x s y G = — 4:a 3 x y 2 , Ans. 



160 ALGEBRA. 



8 X 3 7)1 

3. Find the cube root of 



27 a G b 9 



8 / /8 x 3 m 12 \ _ 2 x m 4 



Find the values of the following : 

* 

4. if - 125 x 3 y 6 . 9. yV" 1 ^. 14. \/ 729 a 1 * b 24 <A 



5. ySla*b 8 . 10. Sl-8an«x\ 15. \/-32 c 6 "^ ™. 

_ J f 32 m 5 n 10 \ . 5/ 

V \ 243 J ' 1L ^ 16 a;2m+2 a2 " 16 - V 243 w 16 »*■ 

7. \/l»^?. 12. y/ ^QQ^l ) • 17. V(« + *)W- 

8. y/ 625 a 12 c 2 . 13. y' 3 2 " 6 3n a n . 18. faj 8B + V 1 " 6 - 

SQUARE ROOT OF POLYNOMIALS. 

239. In Art. 11G we explained a method of extracting the 
square root of a trinomial, provided it was a perfect square. 
We will now give a method of extracting the square root of 
any polynomial which is an exact square. 

Since the square of a + b is a 2 + 2 a b + b 2 , we know that 

the square root of a 2 + 2 a b + b 2 is a + b. If we can discover 

an operation by which we can derive a + b from a 2 + 2 a b + b 2 , 

we can give a rule for the extraction of the square root. 

„ „ 7 7 „ 7 Arranmne the terms of the 

a 2 + 2ab+b 2 a + b 5 °. ' 

2 square according to the descend- 

o „ , 7. 2 a b A- V 1 * n 8 powers of a, we observe thai 

2 a b + b 2 the square root of the first term, 

a 2 , is a, which is the first term 
of the required root. Subtract its square, a 2 , from the uiven 
polynomial, and bring down the remainder, 2 a b + b 2 or 
(2 a + b) b. Dividing the first term of the remainder by 2 a, 
that is, by twice the first term of the root, we obtain b, the 
other term. This, added to 2 a, completes the divisor, 2 a + b ; 



EVOLUTION. 161 

which, multiplied by b, and the product, 2 ab + V 2 , subtracted 
from the remainder, completes the operation. 

By a similar process, a root consisting of more than two 
terms may be found from its square. Thus, by Art. 230, we 
know that (a + b + c) 2 = a 2 + 2 a b + b 2 + 2 a c + 2 b c + c 2 . 
Hence, the square root of a 2 + 2 a b + b' 2 + 2 a c + 2 b c + c 2 is 
a + b + c. 

a 2 + 2ab + b 2 + 2ac+2bc + c 2 

•2 

a 



a + b + c 



2a + b 



2ab + b 2 + 2ac+2bc + c 2 
2ab + b 2 



2 a + 2 b + c 



2ac+2bc + c 2 
2ac+2be + r 2 



The square root of the first term, a 2 , is a, which is the first 
term of the required root. Subtracting a 2 from the given poly- 
nomial, we obtain 2 a b as the first term of the remainder. 
Dividing this by twice the first term of the root, 2 a, we ob- 
tain the second term of the root, b, which, added to 2 a, com- 
pletes the divisor, 2 a + b. Multiplying this divisor by b, and 
subtracting the product, 2 a b + b' 2 , from the first remainder, 
we obtain 2 a c as the first term of the next remainder. 

Doubling the root already found, giving 2 a + 2 b, and di- 
viding the first term of the second remainder, 2 a c, by the first 
term of the result, 2 a, we obtain the last term of the root, c. 
This, added to 2 a + 2 b, completes the divisor, 2 a + 2 b + c ; 
which, multiplied by the last term of the root, c, and subtracted 
from the second remainder, leaves no remainder. 

From these operations we derive the following 

RULE. 

Arrange the terms according to the powers of some letter. 

Find the square root of the first term, write it as the first 
term of the root, and subtract its square from the given poly- 
nomial. 

Divide the first term of the remainder by double the root 
already found, and add the result to the root, and also to the 
divisor. 



162 



ALGEBRA. 



Multiply the divisor as it now stands by the term of the root 
last obtained, and subtract the product from the remainder. 

If there are other terms remaining, continue the operation 
In the same manner as before. 

Note. Since all even roots have the double sign ± (Art. 237), all the 
terms of the result may have their signs changed. In the examples, how- 
ever, we shall consider only the positive sign of the result. 



EXAMPLES. 

1. Find the square root of 9 x* — 12 x s + 16 x 2 — 8 x + 4. 

9a; 4 -12a; 3 + lGx~-8x + 4: 3a; 2 -2a; + 2 

9 a; 4 



6 a: 2 - 2 



./' 



-12 a; 3 
- 12 x 3 



4 a;' 2 



6 x- - 4 x + 2 



12 x 2 - 8 x + 4 
12 a;' 2 -. 8 x + 4 



Ans. 3x 2 — 2x + 2. 
Find the square roots of the following : 



2. ±x i -±x s -3x 2 + 2x + l. 



2 1 

— + — i 
m m 



3. 4 a 4 -16 a 3 +24 a 2 - 16 a + 4. 

4. m 2 + 2 m — 1 

5. 9 — 12 x + 10 :•■- - 4 x 3 + x*. 

6. 19 x 2 + 6 x 3 + 25 + x i + 30 a-. 

7. 28 a- 3 + 4 a; 4 - 14 x + 1 + 45 x 2 . 

8. 40 jc + 25 - 14 x 2 + 9 x* - 24 a; 3 . 

9. 4 .r 4 + 64 - 20 a- 3 - 80 x + 57 x 2 . 

10. a 2 + b 2 + c 2 -2 ab -2 a c + 2 b c 

11. a; 2 + 4 y 2 + 9 « 2 — 4 a; ?/ + 6 a; » — 12 y z. 

No rational binomial is an exact square;; hut, by the rule, 
the (ip/imxliiinte root may be found. 



EVOLUTION. 10- 



• i 



Find, to four terms, the approximate square roots of the fol- 
lowing : 
12. 1 + x. 13. a 2 + b. 14. 1 — 2 x. 15. a 2 + x 2 . 

The square root of a perfect trinomial square may be ob- 
tained by the rule of Art. 116, 

Find the square roots of the first and last terms, and con- 
nect the results by the sign of the second term. 

Extract the square roots of the following : 

16. x 4 + 8x 2 +16. 19. « 2m -4 a m+n + 4 a 2 ". 

rtrt a 2 4 a ±b 2 

17. 9x*-6xf + f. 20. _- — +^. 

t 2 4 x 2 9 ?/ 4 

18. a*-ax + T . 8L 9? + 2 +4^- 

SQUARE ROOT OF NUMBERS. 

240. The method of Art. 239 may be used to extract the 
square roots of numbers. 

The square root of 100 is 10 ; of 10000 is 100 ; of 1000000, 
is 1000 ; and so on. Hence, the square root of a number less 
than 100 is less than 10 ; the square root of a number between 
10000 and 100 is between 100 and 10 ; the square root of a 
number between 1000000 and 10000 is between 1000 and 100 ; 
and so on. 

Or, in other words, the integral part of the square root of a 
number of one or two figures, contains one figure; of a number 
of three or four figures, contains two figures ; of a number of 
five or six figures, contains three figures ; and so on. Hence, 

If a point is placed over evei*y second figure in any integral 
number, beginning with the units' 1 place, the number of point* 
will shoiv the number of figures in the integral part of its 
square root. 



164 ALGEBRA. 



241. Let it be required to find the square root of 4356. 

Pointing the number according to 

60+6 the rule of Art. 240, it appears that 

there are two figures in the integral 



4356 
3600 



120 + 6 756 'J" ° ~~— -~ . 

jgQ part oi the square root. Let a denote 

the figure in the tens' place in the 
root, and b that in the units' place. Then a must be the 
greatest multiple of 10 whose square is less than 4356 ; this 
we find to be 60. Subtracting a 2 , that is, the square of 60, or 
3600, from the given number, we have the remainder 756. 
Dividing this remainder by 2 a, or 120, gives 6, which is the 
value of b. Adding this to 120, multiplying the result by 6, 
and subtracting the product, 756, there is no remainder. 
Therefore we conclude that 60 + 6, or 66, is the required square 
root. 

The zeros being omitted for the sake of brevity, we may ar- 
range the work in the following form : 



4356 
36 



G6 



126 



756 
756 



RULE. 

Separate the given number into periods, by pointing every 
second figure, beginning with the units' [dace. 

Find the greatest square in the left-hand period, and place 
its root on the right ; subtract the square of this root from the 
first period, and to the remainder bring down the next period 
for a dlr hi end. 

Divide this dividend, omitting the last figure, by double the 
root already found, and annex the result to the root and also 
to the divisor, 

Multiply tin' divisorj as it now stands, by the figure of the 
root last obtained, and subtract the product from tin- dividend. 

If there are more periods to be brought down, continue the 
operation in the same manner as before. 



EVOLUTION. 165 

If there be a final remainder, the given number has not an 
exact square root ; and, since the rule applies equally to deci- 
mals, we may continue the operation, by annexing periods of 
zeros to the given number, and thus obtain a decimal part to 
be added to the integral part already found. 

It will be observed that decimals require to be pointed to 
the right ; and if they have no exact root, we may continue 
to form periods of zeros, and obtain decimal figures in the root 
to any desirable extent. 

As the trial divisor is necessarily an incomplete divisor, it is 
sometimes found that after completion it gives a product' larger 
than the dividend. In such a case, the last root figure is too 
large, and one less must be substituted for it. 

The root of a common fraction may be obtained, as in Art. 
238, by taking the root of both numerator and denominator, 
when they are perfect squares. If the denominator only is a 
perfect square, take the approximate square root of the nu- 
merator, and divide it by the square root of the denominator. 
If the denominator is not a perfect square, either reduce the 
fraction to an equivalent fraction whose denominator is a per- 
fect square, or reduce the fraction to a decimal. 

EXAMPLES. 

1. Extract the square root of 49.434961. 



49.434961 
49 



7.031 



1403 



4349 
4209 



14061 



14061 
14061 



Ans. 7.031. 

Here it will be observed that, in consequence of the zero in 
the root, we annex one zero to the trial divisor, 14, and bring 
down to the corresponding dividend another period. 

Extract the square roots of the following : 





ALGEBRA. 






6. 


.9409. 


10. 


.006889. 


7. 


6561 
9025' 


11. 


.0000107584 


8. 


1.170724. 


12. 


811440.64. 


9. 


446.0544. 


13. 


.17015625. 



166 

2. 273529. 

3. 45796. 

4. 106929. 

5. 33.1776. 

Extract the square roots of the following to the fifth decimal 
place : 

14. 2. 16. 31. 18. 

15. 5. 17. 173. 19. 

242. When n + 1 figures of a square root have been ob- 
tained by the ordinary method, n more may be obtained by 
simple division only, supposing 2n + l to be the whole num- 
ber. 

Let N represent the number whose square root is required, 
a the part of the root already obtained, x the rest of the root ; 
then 

y/_ZV= a + x, 
whence, iV= a 2 + 2 a x + x 2 ; 

therefore, JV — a 2 = 2 a x + x 2 , 



7 




1 




20. 




9' 




3' 


3 




2 




21. 




16' 







N-a* 



= x + 



x~ 



2 a '2a 

Then iV— a 2 divided by 2 a will give the rest of the square 

x 
root required, or x, increased by -= — ; and we shall show that 

x 
-= — is a proper fraction, less than \, so that by neglecting the 

Zi a 

remainder arising from the division, we obtain the part re- 
quired. For, x by supposition contains n figures, so that x 2 
cannot contain more than 2 n figures ; but a contains 2n + l 



EVOLUTION. 



167 



figures ; and hence — is a proper fraction. Therefore - 

a 2 a 

is a proper fraction, and less than ^. 

In the demonstration we supposed JV an integer with an 
exact square root ; but the result may be extended to other 
cases. 

From the examples in Art. 241, we observe that each re- 
mainder brought down is the given expression minus the 
square of the root already obtained ; and is therefore in the 
form A 7 " — a 2 . If, then, any remainder be divided by twice 
the root already found, we can obtain by the division as many 
more figures of the root as we already have, less one. 

We will apply these principles to calculating the square root 
of 12 to the. sixth decimal place. We will obtain the first four 
figures of the result by the ordinary method : 



12.000000 
9 



3.464 



64 



300 
256 



686 



4400 
4116 



6024 



28400 

27696 



'04 



The remainder now is .000704 ; and twice the root already 
found is 6.928. Then, by dividing .000704 by 6.928, we can 
obtain the next three figures of the root. Thus, 

6.928). 0007040 (.000102 
.0006928 

11200 

That is, the square root of 12 to the nearest sixth decimal 
place is 3.464102. 

The following rule will be found to save trouble in obtaining 
approximate square roots by this method : 



168 



ALGEBRA. 



Divide the remainder by twice the root already found {omit- 
ting the decimal point), and annex all of the quotient, except 
the decimal point, to the part of the root already found. 

In practice the work would be arranged thus : 



12. 
9 

64 30C 

25C 


3.464 

) 


686 


4400 
4116 


6924 


: 28400 
27696 


6928) 704.000 (.102 
6928 

11200 

Am. 3.464102 



EXAMPLES. 

1. Extract the square root of 11 to the 4th decimal place. 

2. Extract the square root of 3 to the 6th decimal place. 

3. Extract the square root of 61 to the 8th decimal place. 

4. Extract the square root of 131 to the 3d decimal place. 

5. Extract the square root of 781 to the 5th decimal place. 

6. Extract the square root of 12933 to the 4th decimal place. 



CUBE ROOT OF POLYNOMIALS. 

243. Since (a + b) 3 = a s + 3 a 2 b + 3 a b 2 + b 3 , we know 
that the cube root of a 3 + 3 a" b + 3 a b 2 + b 3 is a + b. 



a 3 + 3 a 2 b + 3 a b 2 + b 3 

a" 



a + b 



3a 2 +3ab + b 2 3 a 2 b + 3 a b 2 + b* 
3a 2 b + 3ab 2 + b 5 



EVOLUTION. 1G9 

Arranging the terms of the cube according to the descending 
powers of a, we observe that the cube root of the first term, a 3 , 
is a, which is the first term of the required root. Subtract its 
cube, a 3 , from the given polynomial, and bring down the re- 
mainder, 3 a 2 b + 3 a b' 2 + b 3 or (3 a 2 + 3 a b + b' 2 ) b. Dividing 
the first term of the remainder by 3 a 2 , that is, by three times 
the square of the first term of the root, we obtain b, the other 
term of the root. Adding to the trial divisor 3 a b, that is, 
three times the product of the first term of the root by the last, 
and b' 2 , that is, the square of the last term of the root, completes 
the divisor, 3 a' 2 + 3 a b + b' 2 ; which, multiplied by b, and the 
product, 3 a 2 b + 3 a b' 2 + b 3 , subtracted from the remainder, 
completes the operation. 

If there were more terms, we should proceed with a + b 
exactly as previously with a ; regarding it as one term, and 
dividing the first term of the remainder by three times its 
square ; and so on. Hence, the following 

RULE. 

Arrange the terms according to the powers of some letter. 
Find the cube root of the first term, write It as the first term 
of the root, and subtract its cube from the given polynomial. 

Take three times the square of the root already found for a 
trial divisor, divide the first term of the remainder by it, and 
write the quotient for the next term of the root. 

Add to the trial divisor three times the prod net of the first 
term by the second, and the square of the second term. 

Multiply the complete divisor by the second term of the root, 
and subtract the product from the remainder. 

If there are other terms remaining, consider the root already 
found as one term, and proceed as before. 



EXAMPLES. 

1. Find the cube root of x 6 — 6 x'° + 40 x 3 — 96 x — 64. 



170 

X 

X 


ALGEBRA. 
6 -6a 5 +40a; 3 -96a-64 1 x 2 -2j 

6 


3x i -6x a +4:x i 


—6 a- 5 

— 6 a; 5 +12 a 4 - 8 x s 


3 a 4 - 12 a 3 + 12 a; 2 

-12a 2 +24a; + 16 


-12a; 4 +48a; 3 


3a 4 -12a: 3 + 24a 


+ 16 


- 12 x* +48 x 3 - 96 .r -64 



Ans. x' 2 — 2 a; — 4. 

The formation of the second divisor may be explained thus : 

Regarding the root already obtained, a 2 — 2 a, as one term, 
three times its square gives 3 a 4 — 12 a 3 + 12 x' 2 ; three times 
x~ — 2 a; times — 4 ; gives — 12 x' 2 + 24 x ; and the square of the 
last root term is 16. Adding these results, we have for the 
complete divisor, 3 a 4 — 12 x 3 + 24 x + 16. 

Find the cube roots of the following: 

2. 1 — 6 y + 12 y' 2 - 8 ?/ 3 . 

3. 8 a; 6 + 36 a; 4 + 54 a; 2 + 27. 

4. 64 a; 3 - 144 a b x 2 + 108 a 2 i 2 a: - 27 a 3 6 3 . 

5. a- 6 + 6 a- 5 - 40 a 3 + 96 x — 64. 

6. v/«-l + 57/ 3 -37/ 5 -3y. 

7. a: 3 + 3a;H h-g. 

8. 15 r 4 - 6 a- - 6 a; 5 + 15 x 2 + 1 + a 6 - 20 x*. 

9. a a + 3 a a & + 3a 2 c + 3aJ 2 + 6a6c + 3ac 2 +S 8 + 3 6 a c 

+ 3l> r 2 + c 8 . 

10. 9 a 3 - 21 x 2 - 36 a 5 + 8 a; 6 - 9 x + 42 a- 4 - 1. 

No rational binomial is an exact cube; but, by the rule, 
the approximate root may be found. 



EVOLUTION. 



171 



Find, to four terms, the approximate cube roots of the 
following : 



11. X s + 1. 



12. x s — 



a° 



13. 8 



./• 



o 



CUBE ROOT OF NUMBERS. 

244. The method of Art. 243 may he used to extract the 
cube roots of numbers. 

The cube root of 1000 is 10; of 1000000, is 100; of 
1000000000, is 1000; and so on. Hence, the cube root of a 
number less than 1000 is less than 10 ; the cube root of a num- 



ber between 1000000 and 1000 is between 100 and 10 ; the 
cube, root of a number between 1000000000 and 1000000 is 
betw T een 1000 and 100 ; and so on. 

Or, in other words, the integral part of the cube root of a 
number of one, two, or three figures, contains one figure; of 
a number of four, five, or six figures, contains two figures ; 
of a number of seven, eight, or nine figures, contains three 
figures ; and so on. Hence, 

If a point is placed over every third figure in any integral 
a miller, beginning with the units' place, the number of points 
will show the number of figures lit the integral part of its eube 
root. 

245. Let it be required to find the cube root of 405224. 

Pointing the number according to 
the rule of Art. 244. it appears that 
there are two figures in the integral 
part of the cube root. Let '/ denote 
the figure in the tens' place in the 
root, and b that in the units' place. 



405224 
343000 



70 + 4 



14700 

840 

16 



15556 



62224 



62224 



Then a must be the greatest mul- 



tiple of 10 whose cube is less than 405224 ; this we find to 
be 70. Subtracting a 3 , that is, the cube of 70, or 343000, 
from the given number, we have the remainder 62224. Divid- 
ing this remainder by 3 a' 2 , or 14700, gives 4, which is the 



172 ALGEBRA- 

value of b. Adding to the trial divisor 3 a b, which is 840, 
and b' 2 , which is 16, completes the divisor, 15556. Multiplying 
the result by 4, and subtracting the product, 02224, there is 
no remainder. Therefore we conclude that 70 + 4, or 74, is 
the required cube root. 

The work is usually arranged thus : 



405224 
343 



74 



14700 

840 

16 

15556 



62224 



62224 



RULE. 

Separate the given member into periods, by pointing every 
third figure, beginning ivith the units' place. 

Find the greatest cube in the left-hand period, and place its 
root on the right / subtract the cube of this root from the left- 
hand period, and to the remainder bring down the next period 
for a dividend. 

Divide this dividend, omitting the last two figures, by three 
times the square of the root already found, and annex the quo- 
tient to the root. 

Add together the trial divisor, with two zeros annexed; 
three times the product of the last root figure by the rest of the 
root, ivith one zero annexed ; and the square of the last root 
figure. 

Multiply the divisor, as it now stands, by the figure of the 
root last obtained, and subtract the product from the dividend. 

If there are more periods to be brought down, continue the 
operation in the same manner as before, regarding the root 
already obtained as one term. 

The observations made after the rule for the extraction of 
the square root (Art. 241) are equally applicable to the extrac- 
tion of the cube root. 



EVOLUTION. 



173 



EXAMPLES. 

1 Extract the cube root of 8.144865728. 



8.144865728 
8 


2.012 


120000 

600 

1 

120601 


144865 
120601 




12120300 

12060 

4 


24264728 
24264728 




12132364 








Ans. 2.012. 



Here it will be observed that, in consequence of tbe in 
the root, we annex two additional zeros to the trial divisor, 
1200, and bring down to tbe corresponding dividend another 
period. 



Extract the cube roots of the following : 



2. 1860867. 



4. 1481.544. 



6. 51.478848. 



3. .724150792. 



29791 
681472 



7. .000517781627. 



Extract the cube roots of the following to the third decimal 



place : 



8. 3. 


10. 212. 


12 4 


9. 7. 


11. 5 

8 


13 3 
13 ' 17 



246. JVJien n + 2 figures of a cube root have been obtained 
by the ordinary method, n more may be obtained by division 
only, supposing 2 n + 2 to be the whole number. 



174 ALGEBRA. 

Let N represent the number whose cube root is required, a 
the part of the root already obtained, x the rest of the root ; 
then 

$ 2V= a + x, 

whence, iV= a 3 + 3 a 2 x + 3 a x 2 + x s ; 

therefore, _A r — a 3 = 3 a 2 x + 3 a x 2 + x 3 , 

iV" — a 3 x 2 x 3 



» + — + 



dec a 6 a- 

Then A 7 "— a 3 divided by 3 a 2 will give the rest of the cube 



root required, or x, increased by - 1- 77—,; and we shall show 

a o a 

that the latter is a proper fraction, less than h, so that by 
neglecting the remainder arising from the division, we obtain 
the part required. For, x by supposition contains n figures, 
so that x 2 cannot contain more than 2 n figures. But « con- 

o 

X" 

tains 2 n + 2 figures ; and hence - — is less than ^ . And as 

n — 5= — X o — , and - — is less than 1, - — . must also be less 
o cr a 6 a 6 a -6 a 1 

than T T n . Therefore, (- ^ — ^ is a proper fraction, less than \. 

a k> a 

Remarks similar to those in the last part of Art. 242 apply 
here. 



ANY ROOT OF POLYNOMIALS. 

247. In order to establish a general rule for the extraction 
of roots, it will be necessary to notice the formation of the n\\\ 
power of a polynomial, n being any entire number whatever. 
Thus, 

(a + by = a n + n a n ~ } b + 

Therefore, 

y' a n + n a u ~ l b + = a + b. 

The first term of the root, a, is the nth root of a n , the first 
term of the power; and the .second term of the root, b, may be 



EVOLUTION. 175 

obtained by dividing the second term of the power by n a n ~\ 
or by n times the (n — l)th power of the first term of the root. 

If the root now found be raised to the nth power, and sub- 
tracted from the given polynomial, it will be seen that two 
terms of the required root have been determined. 

It will be observed that the process is essentially that of the 
preceding Articles, simplified by dispensing with completed 
divisors, and generalized. Hence the • 

RULE. 

Arrange the terms according to the powers of some letter. 

Find the required root of the first term for the first term of 
the root, and subtract its power from the given polynomial. 

Divide the first term of the remainder by n times the 
(n — l)th power of this root, for the second term of the root, 
and subtract the nth power of the root now found from the 
given polynomial. 

If other terms of the root require to be determined, use the 
same divisor as before, and proceed in like manner till the nth 
poiver of the root becomes equal to the given polynomial. 

This rule is, also, applicable to numbers, by taking n figures 
in each period. 

EXAMPLES. 

1. Extract tl* cube root of x 6 + 6 x 5 — 40 x s + 96 x — 64. 



x 6 + 6 x 5 - 40 x 3 + 96 x - 64 



(* 2 ) 



2\3 x 6 



x 2 + 2x 



3 a; 4 1 6 x 5 



(x 2 + 2 x) 3 = x 6 +6 x 5 + 12x*+Ssci a 
Sx 4 -12* 4 



(x 2 + 2x-4) 3 = x 6 + 6 x 5 - 40 x 3 + 96 x - 64 

Ans. x- + 2 x — 4. 

2. Extract the cube root of ??i 6 — 6 m 5 + 40 m 3 — 90 m — 64. 

3. Extract the square root of a i —2a 3 x + 3a 2 x 2 —2ax 3 + x i . 



176 ALGEBRA. 

4. Extract the fifth root of 32 x 5 - 80 x* + SO x s — 40 x" 
+ 10 a; — 1. 

5. Extract the fourth root of x s — 4 a 7 + 10 a; 6 — 16a; 5 + 19 x 4 

- 16 a- 3 + 10 x 1 - 4 x + 1. 

248. When the index of the required root is a multiple of 
two or more numbers, we may obtain the root by successive 
extractions of the simpler roots. 

For, since (Art. 17), ( 7 a) mn = a, 
taking the nth. root of both members, we have (Art. 235), 

Taking the mth root of both members, 

y/ a== y (y a). 

Or, the mnth root of a quantity is equal to the mth root of 
the nth root of that quantity. 

EXAMPLES. 

1. Extract the fourth root of 16 x i - 96 x 3 y + 216 x 2 y 2 

- 216 x y s + 81 y\ 

2. Extract the sixth root of a 12 — 6 a 10 + 15 a 8 — 20 a 6 
+ 15 a 4 - 6 a- + 1. 

3. Extract the fourth root of m 8 — 8 m 1 ± 12 w 6 + 40 m 5 

- 74 m* - 120 m s + 108 m 2 + 216 wi + 81. 



XXII. — THE THEORY OF EXPONENTS. 

249. In Art. 17, we defined an exponent as indicating how 
many times a quantity was taken as a factor; thus 

a m means ay. ay, a to m factors. 

Obviously this definition has no meaning unless the expo- 
nent is a positive integer; and as fractional and negative ox- 



EXPONENTS. 177 

portents have not been previously defined, we may give to 
them any definition we please. 

250. We found (Arts. 82, 93, and 228) that when m and 
n were positive integers, 

I. a m Xa n — a m + n . 

a m 
II. — = a m ~ n . 
a n 

III. («'")" = a mn . 

As it is convenient to have all exponents follow the same 
laws, as regards multiplication, division, and involution, we 
shall define fractional and negative exponents in such a way 
as to make Ride I hold for any values of m and n. We shall 
now find what meanings must be assigned to them in con- 
sequence. 

3 

251. To find the meaning of a?. 

As Rule I is to hold universalh T , it follows that 

a a a + a s 
a 2 Xa 2 = a 2 2 = a 2 =a 3 . 

a 
Hence, a 2 is such a quantity as when multiplied by itself 

3 

produces a 3 . Then, by the definition of root (Art. 17), a 2 must 

a 

be the square root of a? ; or, a 2 = \J a z . 

Again, to find the meaning of a 3 . 

i i i. i + i + i a 
By Rule I, a 3 X a, 3 X a 3 ■= a 3 3 3 = a 3 = a. 

Hence, a* is such a quantity as when taken 3 times as a 
factor produces a ; or, a 3 = fya. 

252. We will now consider the general case. 

p 
To find the meaning of a q > p and a being positive integers. 



178 ALGEBRA. 

p_ p p 
By Kule I, a q X a 9 X a 5 X to q factors 

P P P P yy 

— +—-i 1- to? terms — X<2 

= a q q « =a q =a p . 

p 

Hence, a 5 is such a quantity as when taken q times as a 

p 

factor produces a p . Then a' 1 must be the qi\\ root of a p ; 
p 

or, a* = ya p . 

3 4 5 1 

For example, a* = \j a z \ c 2 =\J c 5 ; x 3 ' = y x ; etc., 
and, conversely, y' a 5 = « 4 ; y 7 ic = x 2 ; y" m 5 = m, 3 " ; etc. 

EXAMPLES. 

253. Express the following with radical signs instead of 
fractional exponents : 

a 2 x i 



a 



3.2 c 2 . 5. x*y^. 7. 4a 5 J« 9. 5y T °£ T ^. 



2.6^. 4. 3«m^. 6, rn'iA 8. 2c«t/l 10. 3»^c^l 

Express the following with fractional exponents instead of 
radical signs : 

11. yV. 13. sjn. 15. 3 y/ m 5 . 17. ty a 1 ty a?. 

12. yV- 14. yV. 16. 4 ^a 10 . 18. v^vV- 

19. 5sjm»%n<. 20. 2 av Vy>. 

254. To find the meaning of a -3 . 
By Kule I, a~ 3 X a 3 = a = 1, by Art. 94. 

Hence, a ~ 3 = — - . 

To find the meaning of a~ 2 . 
By Rule I, a~* X c$ — a = 1. 

Hence, a. 2 = — . 



EXPONENTS. 179 

255. "We will now consider the general case. 

To find the meaning of a~ s , s being integral or fractional. 

By Eule I, a~ s X a s = a° = 1. 

1 

Hence, a s = — . 



1 1 -2 1 

For example, ar 1 = ~\ a 4 — ^?5 a 3 — ~ '■> e * c v 

a' 



a a* 



1 X 2 2-3 

and, conversely, — ^ = « -2 ; — = a; 2 a -3 ; — = 2 a * ; etc. 

We observe, in this connection, the following important 
principle : 

^4 quantity may be changed from the denominator of a frac- 
tion to the numerator, or from the numerator to the denomi- 
nator, if the sign of its exponent be changed. 

EXAMPLES. 

256. Remove all powers from the denominators to the 
numerators in the following : 



ar 5a; 3 2 a; -1 a; a; 2 , a; -2 x~ 

a 2 a 3 — 1 a 4 a 5 — b 



3. 



4. 



x 3 


3 

a- 4 


+ 


a; 2 a; 5 


7 m 
6c- 1 


3 m 

3 

7 c* 




4m 2 -l 3??? 3 + 2?i 

i + 
5 c 6 2 c" 5 



Remove all powers from the numerators to the denominators 
in the following : 



*o 



_ 2a; 3a" 2 a _ a? a: 3 x~ 2 2x 1 



180 ALGEBRA. 



a 6 3a 4 5a 2 a 

7. —^--^ — ^ + 



8. 



a; + 2 5 b 03 7 — b a' 



m — x m 3 % 5 2p 



1 — x 2 3x 5x 1 7 x 



K ™ — 1 7 ^. — 3 * 



Express the following with positive exponents: 

J_ 2. _3 

9. 2x 2 y 2 — 3x~ l y i '— x~ i y r . 

10. a- 1 5- 2 + 2cr 3 i- 4 -3fth~l 

_i _s _i 

11. 3a; 3 y 7 — 4 a* ?/ + a y~ • 

12. a" 1 6~ 2 c 8 + a~ 2 b~% e -4 + a s j-2 c# 

257. We obtained the meanings of fractional and negative 
exponents on the supposition that Bule I, Art. 250, was to 
hold universally. Hence, for any values of m and n, 

a m x a n — a m + n 

3 2 3_2. J^ 

For example, a 2 X «~ 5 = a 2 ~ 5 = a~ 3 ; a 4 X « — «* 3 = a ' 5 
a- i Xa* = a 2 = a 2 ; a 3 X a 5 = a 3 ° = a To I etc. 



EXAMPLES. 

Multiply together the following : 

1. a 8 and a" 1 . 4. c 8 and y'c 2 . 7. rc and n~?. 

2. a 2 and a" 2 . 5. a:" 1 and (far 8 . 8. a^ and af*. 

3. a- x and a~ 5 . 6. m 2 and ^— . 9. 2 c~ * and - 3 a tyc s . 



EXPONENTS. 181 



10. Multiply A ^+2a*-3ftHy26 *-4a 3_6a £&*. 



a% b~ 2 + 2 a^ - 3 b 2 
2 J~2 _ 4 a -J _ o a ~% iih 

2 a$ &- 1 + 4 a* 6"^ - 6 



Aa*b~*-S + 12a H 2 . 



6 - 12 a - J b 2 + 18 a %b 



2 a$ b- 1 - 20 + 18 a » 6, ^ws. 

Note. It should be carefully remembered, in performing examples like 
tbe above, that any quantity whose exponent is is equal to 1 (Art. 94). 

Multiply together the following : 

11. a 2 b~°- - 2 + a~ 2 b 2 and a 2 b~ 2 + 2 + a~ 2 b 2 . 

3. i i ii a , i i 

12. a 4 — a? b± + a 4 b' 2 - b* and a 4 + b*. 

13. a~ 2 - 2 a" 1 b + b 2 - a b 3 and ar s + 2 or 2 b. 

14. 3 ar 1 -a~ 2 b- 1 + or* b~ 2 and 6 a* b 2 + 2 a 2 b + 2a. 

15. x~ s f -x~ 2 y-2x- 1 and 2 x 2 y~ l + 2x 8 ! /- 2 -±x i y~\ 

16. x% y~* + 2 + jc" * y* and 2 a; - * y 4 _ 4 A .- $ y f + 2 a-- 2 y 4 . 

17. 2 sc^ — 3 a;^ — 4 + aT* and 3 a;'*' + x — 2 x*. 

18. 4 a 4 &- 1 + « 4 - 3 «~ 4 6 and 8 a 4 &- 1 - 2 a~± - 6 «~ 4 6. 

258. To prove that Rule II holds for all values of 'm and n. 
By Rule I, a m ~ n X a n = a m ~ n + w = a m . 

Inverting the equation, and dividing by a n , we have 



— = a" 
a n 



182 ALGEBRA. 



ft 3 cc~ 2 

For example, — =:a 3_1 =:a 2 ; — — = a~ 2 ~ 3 = a~ 5 ; 



-I _| + 2 f a 3 8 + 4 V- 



a 



- = a 4 ' " = a 4 ; — - = a 5 = a 5 ; etc. 



EXAMPLES. 

Divide the following: 

_i _4 l 1 

1. a 3 by a -1 . 4. a 2 by a . 7. x by -^— g. 

2. a by a 3 . 5. rMiyJc 5 . 8. 5 » by 2 or 1 $ 6. 

3. a^ by a* 6. m 2 by tym,- 2 . 9. a" 1 6^ by - 3 a 6~~*. 

10. Divide 2 a^ 6" 1 -20 + 18 a - """ 6 by J b~ * + 2 J - 3 6*. 



2 J' b- 1 - 20 + IS a ^b 
2a?b~ 1 +4 : cfib~^—6 



2 _l JL I 

a :i 6 »+-2a* — 3 6' 



2 6 2 - 4 a * — 6 a $ b 2 , ^?*s. 



-4 a* 6 * — 14 + 18 a ^6 
-4a^6~^- 8 + 12 »~^ 6^ 

-6 -12 a"* 6^+ 18 a~ %b 

- 6 - 12 a~^ 62 + 18 <T% b 



Note 1. Particular attention must be given to seeing that the dividend 
and divisor are arranged in the same order of powers, and that each re- 
mainder is brought down in the same order. It must be remembered that 
a zero exponent is greater than any negative exponent ; and that negative 
exponents are the smaller, the greater their absolute value. 

Note 2. In dividing the first term of the dividend or remainder by the 
first term of the divisor, it will be found convenient to write the quotient 
at first in a fractional form; reducing the result by the principles of Art. 
258. Thus, in getting the first term of the quotient in Ex. 10, we divide 

2 a* b- 1 by a? b ' 2 . Then, the result = „ _ L = 2 a 3 3 b + * = 2 b * 

(fit * 



EXPONENTS. 183 

Divide the following : 

jl i 

11. a — b by a 5 — b b . 

12. a- 4 + a~ 2 b' 2 + b~ i by a~ 2 + a' 1 b' 1 + b~ 2 . 

13. 2x- 2 y 2 + 6 + Sx 2 i/- 2 hj2x + 2x 2 y- 1 + 4 t x 3 y- 2 . 

14. 2 x ?s y~ x — 2x~%y + 32 x~ 2 y 3 by 2 + 6 a-" § y + 8 as" * ?/. 

15. cc~ 3 ?/ -5 — 3 x~ 5 y~ n + x -1 y~ 9 by x~ 2 y~ s + x~ 3 ?/ -4 — a; - 4 y -6 . 

16. 8 — 10o;- 2 2/ J ^+2aj-*2/^ > " by 4x~*y% + 2x~ 2 y*—2x~^y 4 . 

259. To prove that Rule III holds for all values of m 
and n. 

We will consider three cases. 

Case I. Let m have any value, and n be a positive in- 
teger. 

Then, from the definition of a positive integral exponent, 

(a m ) n = a m Xa m Xa m to n factors 

— f,m + m + m ton terma __ ~m n 

Case II. Let m have any value, and n be a positive frac- 

v 
tion, which we will denote bv - • 



P_ q, 



Then, (a m ) n = (a m )T= \ (a m ) p , by the definition of Art. 252, 

= y 7 ^"^ by Case I, Art. 259, 



mp 

= aJ, by Art, 252, 



= a mx q = a mn . 



Case III. Let m have any value, and n be a negative 
quantity, integral or fractional, which we will denote by — s. 



184 ALGEBRA. 

Then, (a m ) n = (a m )~ s — — -r s , by the definition of Art. 255, 

(a m ) s 

= , by Cases I and II, Art. 259, 



= a - ms = a m{ - s) = a mn . 

Thus, we have proved Kule III to hold for all values of m 
and n. 

For example, (a 2 ) 3 = a 6 ; (a" 1 ) 5 = a" 5 ; (a~ ?s ) * = a"* ; 

(J)$ = a; (a*)~* = a ~*J (a 2 )~* = a-*; etc. 



EXAMPLES. 

260. Find the values of the following : 

5 '( C " *)">. 10. f ] 



1. («•)-■ 4. (O - *- 7. tf( c 2 ) 2 . 10. - . 



2. («- 2 ) 3 . 5. (e-*)- 2 *. 8. tfm 8 ) * 11. (^ 

3. (a 3 )*. 6. (,/*)* 9. (^f)- 5 . 12. {(^VY 1 - 

261. To prove that (a £) n = a n b n for any value of n. 

In Art. 228 we showed the truth of the theorem when n was 
a positive integer. 

Case I. Let n be a positive fraction, which we will denote 

v E E E 

by — . We have then to show that (ab)« =a« b'i. 
9. 

[(a bfy = (a b) p , by Art. 259. 



p p 



EXPONENTS. 185 

[at &*]* = (c£y (b^) q = a p bP= (ab)", by Art. 228. 

Hence, [( a J)fj*=[a? J?] 3 . 



Therefore, (a fi) «= a« &«. 

Case II. Let w be a negative quantity, which we will de- 
note by — s. We have then to show that (a b)~ s = a~ 3 b~ 3 . 

(a b)-° = -y—^- = — ; - , by Art. 228 and Case I, 
v J (a b) s a s b s J 

= ar s b~ s . 



262. To find the value of a numerical quantity affected 
with a fractional exponent. 

1. Find the value of 8*. 

From Art. 252, we should have S^^y'S 5 ; and to find the 
value in this way, we should raise 8 to the fifth power, and 
take the cube root of the result. 

A better method, however, is as follows : 

8^ = (8*) 5 , by Art. 259, 
= ($&)* = 2 s = 32; Ans. 

Note. Place the numerator of the fractional exponent as the exponent 
of the parenthesis, and 1 divided by the denominator as the exponent of the 
cpiantity within. 

2. Find the value of 16"*. 



| = J__ J_ J_ J_ 1_ 

~ 16* ~ (IB*)* = ^ 16 ) 5 ~ (± 2 > 5 ~ ±32 : 



186 



ALGEBRA. 



EXAMPLES. 

Find the values of the following : 

3.27?. 5.1000-* 7. (-8)1. 9. ('f^f . 

36' 2 x 16 * 

1 7 s 4^ x 9~ 2 

4. 36*. 6. 9"* 8. (-27)* 10. -^T 1 



81" X 16* 



If the numerical quantity is not a perfect power of the de- 
gree indicated by the denominator of the fractional exponent, 
the first method explained in Ex. 1, Art. 262, is the best. 

For example, to find the value of 7 2 , we write it \/ 7 s , or 
y/ 343 ; and find the square root of 343 to any desired degree 
of accuracy. 

MISCELLANEOUS EXAMPLES. 
263. Extract the square roots of the following : 

4 x y c*de 2 

5. 9 x-* y 2 -12x~ s y-2 x~ 2 + 4 x~ x y~ x + y~\ 

6. 4x* + ±x% y~* -15x 2 y~* -Sx% y~* + lGx§ y~\ 

7. z 8 y~% + 6 - 4 x~% y% + x~ 3 y% - 4 x% y~*. 



V 
Extract the cube roots of the following : 

8. ab\ 9. -8a;- 4 / 10. 

11. $ if - 12 y^'x-^ 6 y$x- 2 -i/x- 3 . 



3m 2 n * 



ax 5 



EXPONENTS. 187 

Reduce the following to their simplest forms : 






12. ,^„„^. . 15. a x ~ v+2z a 2x+v ~ 3z a z . 



Ji + 2»i+r 



13. (x a )- b +(x- a )- b . 16. (^'"x^X^- 1 ) *. 

ia (a x+v Y ( a? \ x ~ v ,_ r/ _J_\„_»-i_2_ 

Change the following to the form of entire quantities : 
18 15 ^* 2 19 X *V 2 20 ^ 2 



Reduce the following to their simplest forms : 



21. ' ^ 22. ^=^. 23 



Factor the following expressions : 
24. 9^-12^ + 4. 25. a ^-3a*-88. 



26. ar 2 & + 5 a" 1 &*- 66. 



Factor by the method of Art. 117 : 

27. a-b. 28. a£-&~£. 29. ar'"y — 4«*. 

Factor by the method of Art. 119 : 
30. a — 6. 31. a + 6. 32. x~ s +8cm^. 



188 ALGEBRA. 



XXIII. — RADICALS. 

264. A Radical is a root of a quantity, indicated by a 
radical sign ; as, yfa, \Jx + 1, y m 2 — 2 n + 3. 

When the root indicated can be exactly obtained, it is called 
a rational quantity ; and when it cannot be exactly obtained, 
it is called an irrational or surd quantity. 

265. The Degree of a radical is denoted by the index of 

© / 

the radical sign; thus, \] a is of the second degree; \x + 1 
of the third degree. 

Similar Radicals are those of the same degree, with the 

5/- 6, 

same quantity under the radical sign ; as, \ax and 1 y ax. 

266. Most problems in radicals depend for their solution 
on the following important principle : 

For any values of n, a, and b, by Art. 236, 



V«X \b = \ab. 



REDUCTION OF RADICALS. 

TO REDUCE RADICALS OF DIFFERENT DEGREES TO EQUIVALENT 
RADICALS OF THE SAME DEGREE. 

267. 1. Reduce y 1 2, ^3, and ^ 5 to equivalent radicals of 
the same degree. 

By Art. 252, ^2 = 2* = 2& =^2" = ^64 



{f 3 = 3^ = 3^ = v^ 4 = y'Sl 
^5 = 5* = 5& = y' 5 3 = v' 125 



RADICALS. 189 



RULE. 



Express the radicals 10 it h fractional exponents ; reduce these 
fractions to a common denominator • express the resulting 
fractional exponents with radical signs; and, finally, reduce 
the quantities under the radical signs to their simplest forms. 

Note. This method affords a means of comparison of the relative mag- 
nitudes of two or more radicals ; thus, in Example 1, as y/ 125 is evidently 
greater than y/Sl, and y'Sl than y/64, hence A'o is greater than y/3, and 
^3 than y/2. 

EXAMPLES. 

Reduce the following to equivalent radicals of the same 
degree : 

2. s/3, \f4, and ^5. 5. \f2~^ \J3~b, and ^4^ 

94 ** 6, 4. 

3. y 5, y 6, and y 7. 6. y a + b and y a — b. 

.3/ -, .*/ „ . /— 5 ; -■ 3/- 



4. SJxy, y x z, and yyz. 7. y'cr — x 2 and sj a z — X s . 

8. Which is the greater, ^3 or y'S? 

9. Which is the greater, </2 or y/3? 
10. Which is the greater, ^4 or $5 ? 

TO REDUCE RADICALS TO THEIR SIMPLEST FORMS. 

268. A radical is in its simplest form when the quantity 
under the radical sign is not a perfect power of the degree 
denoted by any factor of the index of the radical, and has no 
factor which is a perfect power of the same degree as the 
radical. 

CASE I. 

269. When the quantity under the radical sign is a perfect 
power of the degree denoted by some factor of the index of the 
radical. 



190 ALGEBRA. 

a 

1. Reduce y 8 to its simplest form. 

EXAMPLES. 

Reduce the following to their simplest forms : 



2. #9. 4. ^27. 6. \/~a^W. 



3. ^25 a 2 . 5. $125 a 3 1> 9 . 7. <l 25 "' 



CASE II. 



270. When the quantity under the radical sign has a 
factor which is a perfect power of the same degree as the 
radical. 

1. Reduce \J 32 to its simplest form. 



V 32 = V 16x2 = (Art. 266) ^16x^2 = 4^2, Am. 



2. Reduce y/54 a 4 x to its simplest form. 

\/5±a i x = \/2Ta s x2ax = % 27~a~ 3 X ^2~a~x = 3 a \/2~a~x~, 

Ans. 

RULE. 

Resolve the quantity under the radical sign into two factors, 
one of which is the greatest perfect power of the same degree 
us the radical. Extract the required root of this factor, and 
prefix the result to the indicated root of the other. 

Note. In case the greatest perfect power in the numerical part of the 
quantity cannot be readily determined by inspection, it may always be ob- 
tained by resolving the numerical quantity into its prime factors. Let it 
be required, for example, to reduce ^ 1944 to its simplest form. 1944 = 
2x2x2x3x3x3x3x3 = 2 8 x3 5 . Hence, 

^1944 - V / 23lT35 = V2 2 x 3 4 x y/6 = 18 ^6. 



RADICALS. 191 



EXAMPLES. 

Reduce the following to their simplest forms 



3. ^50. 6. ^320. 9. 7^63aH 5 c 6 . 

4. 3^24. 7. 2^80. 10. %250x 3 fz». 

5. si 12. 8. \j98a s b 2 . 11. <Jl8x 3 y i -27 x* y 



12. \/ax 2 — 6ax + 9a. 14. \/20 a x 1 + 60 a 2 x + 45 a 3 . 

13. SJix^-y 2 ) (x + y). 15. ^192 a 4 6 5 + 320 a 3 6 4 . 

When the quantity under the radical sign is a fraction, mul- 
tiply both terms by such a quantity as will make the denomi- 
nator a perfect power of the same degree as the radical. Then 
proceed as before. 

/2 

16. Reduce t / - to its simplest form. 

V / i=\/!=V / (H=Y / i x Y /6 =^ G '- te - 

/9 

17. Reduce t / - to its simplest form. 

Reduce the following to their simplest forms : 
18. 






19 
20 



192 ALGEBRA. 

TO REDUCE A RATIONAL QUANTITY TO A RADICAL FORM. 
271. 1. Eeduce 3 a; 2 to a radical of the third degree. 



3 x 2 = \7 (3 x 2 ) 3 = \]21 x\ Ans. 

RULE. 

Raise the quantity to the power indicated by the given root, 
and write it under the corresponding radical sign. 

EXAMPLES. 

Reduce the following to radicals of the second degree : 
a rr a 3x x — 3 

2. i a. 3. -=- . 4. a + 2 x. 5. . 

5 x — 2 

6. Reduce -=- to a radical of the fourth degree, 
o 

TO INTRODUCE THE COEFFICIENT OF A RADICAL UNDER THE 

RADICAL SIGN. 



272. 1. Introduce the coefficient of 2 a y 3 x 2 under the 



radical sign. 



2 a V 3a- 2 = \/ 8 a 3 X V 3 x 2 = (Art. 266) \f 8 a 3 X 3 x 2 = y 7 24 a 3 x% 

Ans. 

RULE. 

Reduce the coefficient to the form of a radical of the given 
degree; multiply together the quantities under the radical 
signs, and write the product under the given radical sign. 

EXAMPLES. 

Introduce the coefficients of the following under the radical 
signs : 

2. 3^5- 4. 4a 2 \/^U. 6. 5c$Ja. 

3. 2^7. 5. 3yTT^. 7. ( x -l)J(?±^). 



RADICALS. 193 

ADDITION AND SUBTRACTION OF RADICALS. 

273. 1. Find the sum of y/18, \J21, J -, and 12 y/^ 



18 



By Art. 270, . ^18 = 3^2 

v/27= 3^3 



12 









2. Subtract ^48 from ^162. 

By Art. 270, ^162 = 3^6 

^48 =2^6 



^6, ^ws. 



RULE. 



Reduce each radical to its simplest form. Combine the 
similar radicals, and indicate the addition or subtraction of 
the dissimilar. 



EXAMPLES. 

Add together the following radicals : 

5 



3. y'S, ^18, and y/50. 6. ^20, t/i and J 

4. ^12, v/48, and v/ 108. 7. y/|y/|> and y^ 

5. ^16,^/54, and v' 128. 8. t/i'tf^**^ 



2 

27 



7 2 

3 



194 ALGEBRA. 

Subtract the following : 

9. v/ 45 from ^ 245. 10. J ~ horn J 

Simplify the following : 



16 
15 



11. \/243 a b 2 + V 75 a* +^3a s - 54 a 2 b + 243 a b 2 . 

12. 7^27-^75-y/| + v/12-y/l-y/l. 

13. {^16 + 5^54-^250-^/^ + ^/1 + ^/^. . 



15. V 63 « 2 « -S4:abx + 2Sb 2 x — ^7d 2 x + 42abx + 63b 2 x. 



MULTIPLICATION OF RADICALS. 

274. 1. Multiply sj 2 by ^ 5. 



^2x^5 = (Art. 266) ^2 X 5 = ^10, Arts. 

2. Multiply y' 2 by y' 3. 

Reducing to equivalent radicals of the same degree (Art. 
267), we have 

y/2 X $3 = $8 X \j 9 = ^72, Am. 



KULE. 

Reduce the radicals, if necessary, to equivalent ones of the 
same degree. Multiply together the quantities under the radi- 
cal signs, and write the product under the common radical 
sign. 



RADICALS. 195 



EXAMPLES. 

Multiply together the following : 



3. y/ 12 and y/ 3. 6. V 6 a* and \/ 5 a?. 

4. ^2 and y^. 7. V^i V^ 4 , and V^^ 6 ) ' 

5. ^axand^bx. 8. y/ 2, y/ 5, and t/ ^ • 



9. Multiply 2y/cc — 3y/?/ by 4y/a: + y/?/. 

2 y/cc — 3 \J y 
4y/a;+ y/y 



8ic — 12 \jxy 

+ 2v/^7-3 2/ 



8 « — 10 v/a; y — 3 ?/, ^4ras. 

Note. It should be remembered that to multiply a radical of the second 
degree by itself is simply to remove the radical sigu ; thus, 

y/ x x y/ x = x. 
Multiply together the following : 
10. s/x- 2 and y]x + 3. 11. 3y/z - 5 and 7 sjx- 1. 



12. s/.T + l-V/z-land y'a; + 1 + y/x-1 (Art. 106). 



13. yV — 1 — a and V «' 2 — 1 + «• 

14. \J x — s] y+ \J z and \j x -\- \J y — \/ z. 

15. y/2-y/3 + y/5 and ^2 + ^3 + ^5. 

16. 3y/5-2y/6 + y/7 and 6 y/5 + 4 y/6 + 2 y/7. 

17. 4y/3-5y/2-2y/5 and 8 y/3 + 10 y/2 - 4 y/5. 



196 ALGEBRA. 

Simplify the following : 

Square the following (Arts. 104 and 105) : 



20. 2^3-^/2. 22. \/l-a 2 +a. 

21. 3^8 + 5^3. 23. ^^b-\{a~+~b. 

DIVISION OF RADICALS. 



275. Since (Art, 266), \faXs/b = \fab, it follows that 

\a b -i-tya = y / 6. 



RULE. 



Reduce the radicals, if necessary, to equivalent ones of the 
same degree. Divide the quantities under the radical signs, 
and write the quotient under the common radical sign. 



EXAMPLES. 

8/ 



1. Divide ^ 15 by y/ 5. 

Reducing to equivalent radicals of the same degree, we have 

^15-^5 = ^225-^125 = ^/11 = ^/1, Ans. 

Divide the following : 

2. v' 108 by v' 18. 6. s/2by$/3. 

3. V^O^by \l~2Z. ■ 7, % 2 by ^3. 

4. v/54b yv /6. 8. s/ 12 hjsj 2. 

5. S/lT^by $3~o~. 9. \/Ta~by$Ta~. 



RADICALS. 197 



INVOLUTION OF RADICALS. 

276. 1. Raise fy 2 to the fourth power. 

(^ 2) 4 = (2*) 4 = 2* = f 2 4 = ^ 16, Jn». 
2. Raise y/ 3 to the third power. 

(ft 3) 3 = (3") 3 = 3^ = 3* = \J 3, Ans. 

We observe that in the first example the quantity under the 
radical sign is raised to the required power ; while in the sec- 
ond, the index of the radical is divided by the exponent of 
the required power. Hence the following 

RULE. 

If possible, divide the index of the radical by the exponent 
of the required poioer. Otherwise, raise the quantity tinder 
the radical sign to the required poioer. 

Note. If the radical has a coefficient, it may be involved separately. 
The final result should be reduced to its simplest form. 

EXAMPLES. 

3. Raise y/5 to the third power. 



4. Square \J1. 

5. Find the fourth power of 4 y 3 x. 

6. Find the sixth power of y/ a 2 x. 



7. Raise \l a — b to the fourth power. 

8. Raise 3 a\bx to the fourth power. 

9. Find the value of (\/ x + l)\ 



10. Find the square of 4 V^ 2 — 3. 



198 ALGEBRA. 

EVOLUTION OF RADICALS. 



277. 1. Extract the square root of y 6 x 2 . 
\J( y^) = (y 7 ^ 2 )* = {(6 x*)*} * = (6 a; 2 )* = \/~6x~% Ans. 



2. Extract the cube root of \/27 x 3 . 

V (V27^Q = (V27V)*= {v/(3 z) 3 P~ = {(3 a>)*}* = ( 3 a ')" 
= y 3 £, ^4?zs. 

We observe that in the first example the index of the radical 
is multiplied by the index of the required root ; while in the 
second, the required root is taken of the quantity under the 
radical sign. Hence the following 

RULE. 

If possible, extract the required root of the quantity under 
the radical sign. Otherwise, multiply the index of the radical 
by the index of the required root. 

Note. If the radical has a coefficient, which is not a perfect power of 
the same degree as the required root, it should he introduced under the 
radical sign hefore applying the rule. Thus, 



y(i\lax ) = y(\?l6ax ) = ylGax. 
The final result should be reduced to its simplest form. 

EXAMPLES. 

3. Extract the square root of y/2. 

4. Find the cube root of y/ 8. 

4 

5. Find the cube root of \a + b. 



6. Find the square root of \x 2 — 2 x + 1. 

7. Extract the fifth root of y/32. 



8. Extract the cuhe root of y/27. 

9. Find the value of ^(3 y/3). 



RADICALS. 199 

5/ 



10. Find the fourth root of \ x* y vi . 

11. Find the value of \J(±\J2). 

TO REDUCE A FRACTION HAVING AN IRRATIONAL 
DENOMINATOR TO AN EQUIVALENT ONE 
WHOSE DENOMINATOR IS RATIONAL. 

CASE I. 

278. When the denominator is a monomial. 

2 b 
1. Reduce -; — to an equivalent fraction whose denominator 

sj a 

is rational. 

Multiplying both terms by y/ a, 

2b _2bsja _2bsja 



\l a y/ a\j a a 



, Ans. 



5 
2. Reduce ^-^ to an equivalent fraction whose denominator 

is rational. 

Multiplying both terms by y/ 9, 

5_ 5^9 _5v^9_5^9 



RULE. 

Multiply both terms of the fraction by a radical of the same 
degree as the denominator, with such a quantity under the 
radical sign as will make the denominator of the resulting 
fraction rational. 



200 ALGEBEA. 



EXAMPLES. 



Reduce the following to equivalent fractions with rational 
denominators : 



3 3 
3< ^2" 


4 1 
$2 a 


5 5 
5 ' IN- 
CASE II. 


6- r 2c 
V '27 a 2 



279. When the denominator is a binomial, containing only 
radicals of the second degree. 

1. Reduce — — - r^ to an equivalent fraction whose denomi- 
nator is rational. 

Multiplying both terms by 3 — y/ 2, 

10 10( 3-^2) fK ^ iA ^ 30-10y/2 
3T72 = (3 + V2)(3-^2) = ^ Art 106 > 7 ' ^ 

2. Reduce — —- to an equivalent fraction whose de- 
nominator is rational. 

Multiplying both terms by ^ 5 + \J 2, 

v/5 + v /2 _ ( v /5 + v/2)(y/5 + v/2) _ 7 + 2 v /10 
v/5-v/2 _ ( v /5- s /2)( v /5 + v /2)~" 3 ; AnS ' 

RULE. 

Multiply both terms of the fraction by the denominator with 
the sign between its terms changed. 

EXAMPLES. 

Reduce the following to equivalent fractions with rational 
denominators : 



RADICALS. 201 



4 2y/5 + y/2 s/a + x+^a-x 

3< 3TV2* " s/5-3^2' • y/^r^-y^T 



a; 



, 4-y/3 y/«-y^ 19 yV-l-y/^+1 

2-v/3 s/a+s/x ^at-l+^at+l 

_ VS-V 73 o 2 +V« + 1 1Q ^ + V/^ 2 -4 

«• ~777 To • "• , • A«5- , • 

V/2 + v/3 1-Va+l » — VaJ 2 -4 



y/ft + y/5 1Q « — V/V— 1 14 y/x — 4 y/a — 2 

sja-\jb' ' a + yjd 1 —! ' 2\lx + 3\jx — 2* 

280. If the denominator is a trinomial, containing only 
radicals of the second degree, by multiplying both terms of 
the fraction by the denominator with one of its signs changed, 
we shall obtain a fraction which can be reduced to an equiva- 
lent fraction with a rational denominator by the method of 
Case II. Thus, to reduce the fraction 

y/2-v/3-v/7 
^2 + ^3 + ^' 

Multiplying both terms byy/2 + y/3 — y/7, 

v /2- v /3-v/7 _ (v/2-v/3-v/7)(v/2 + v /3-v/T) _6-2 v 14 
l/2 + y/3 + y/7~(\/2 + s/3 + \/7)(s/2+sJ3-s/7) 2y/6-2 

_ 3-y/14 

Multiplying both terms by y/ 6 + 1, we Lave 

(3- V /14)( v /6 + l) _ 3-v/14 + 3y/6- v /84 
(v / 6 _l )(v /6 + l) - 5 

If the denominator is a binomial, containing radicals of any 
degrees whatever, it is possible to reduce the fraction to an 
equivalent form with a rational denominator ; but the process 
is more complicated than the preceding and rarely necessary. 



202 ALGEBEA. 

281. To find the approximate value of a fraction whose 
denominator is irrational, reduce it to an equivalent fraction 
whose denominator is rational. 

1. Find the value of ^ j^ to three decimal places. 

2 -J^ = (Art. 279) 2 -±^- 2 = 2 -±|^ = 1.707, An*. 

It will be seen that the value of the fraction is obtained in 
this way more easily than by dividing 1 by 2 — \J 2, or its 
value .586. 

EXAMPLES. 

Find the values to three decimal places of the following : 

2 2 3 3 4 - 5 V /3 -V /2 



IMAGINARY QUANTITIES. 

282. An Imaginary Quantity is an indicated even root of 

a negative quantity ; as, y — 4, \/ — a 2 . 

In contradistinction, all other quantities, rational or irra- 
tional, are called real quantities. 

283. All imaginary quantities may be expressed in one 

common form, which is, a real quantity multiplied by y — 1. 
For example, 

y/3^ _ yV X (_i) _ (Art. 266) y' a" X ^ :Z 1 = a y^l ; 



also, y/- 2 = ^2 X(-l) = s/2\[^l. 

Hence, we may regard ^—1 as a universal factor of every 
imaginary quantity, and use it in our investigations as the 
only symbol of such a quantity. 



KADICALS. 203 

284. Imaginary quantities may be added, subtracted, and 
divided the same as other radicals ; but with regard to multi- 
plication, the usual rule requires some modification. 



285. By Art. 17, V— 1 means such an expression as when 
multiplied by itself produces — 1 ; 

or,- (v/^l7 = -l; 

also, (y/I^^v/^lTxV^T^-lV 3 !; 

and, (V^1) 4 = (V^1) 2 X(V^ 3 1) 2 =(-1)X(-1) = 1. 

By continuing the multiplication, we should find 

(V-l) 8 = l; etc. 
Or, in general, where n is any positive integer, 

(v/=ir + W=T; (V=i) 4B+2 =-i; (V=T) 4n+8 =-V^; 

(V=1) 4 " +4 = L 

MULTIPLICATION OF IMAGINARY QUANTITIES. 

286. 1. Multiply ^^a^hy \f^¥. 

V^^X V 17 ^^ (Art. 283) a <f^l X ft y'^1 = « ft (V 11 *) 2 

= — aft, ^4hs. 



2. Multiply V- 2 by y^3. 

V^2 X y /r 3=v/2x\/3x(v /:: i) 2 =-^ ) il 



??s. 



3. Multiply together ^— 4, V— 9, V~ 16 > and V" 25. 

V^ X V 3 ^ X S/^IG x V /Ir 25 = 2x3x4x5x (v^)* 
= 120(\/~l) 4 = 120, Arts. 



204 ALGEBRA. 

RULE. 

Reduce all the imaginary quantities to the form of a real 
quantity multiplied by ^— 1. Multiply toy ether the real 
quantities, and multiply the result by the required power of 

EXAMPLES. 

Multiply the following : 



4. 4^-3 and 2\J-2. 7. 1 + V- 1 and 1 - \J- 1. 



5. _ 3 yL_ a and 4 V- b. 8. \J- a% V~ b% and y— c 2 . 



6. 4 + V- 7 and 8-2^-7. 9. a + \/— b and a — \J—b. 



10. 2 V- 3 — 3 V- 2 and 4 V- 3 + 6 V~ 2. 



11. Divide V- « by V~ *• 



We should obtain the same result by using the rule of Art. 
275 ; hence, that rule applies to the division of all radicals, 
whether real or imaginary. 

Divide the following : 



12. V- 6 by V- 2. 

13. ^^12 by y'-S. 

Simplify the following : 



16. 1 + S ^_1 . (Art. 279). 

1-V-l 2-V-2 



14. 


V- 


-5hj^- 


1. 


15. 


tf- 


- 54 by y 7 - 






17. A + t 


-2 



18. ^£0 + "-^ . (Art. 154). 
a — \'—b a +^-b 

19. Expand (2 -V^3) a . 20. Expand (2 + 3 Y^2) 8 . 



KADICALS. 205 

QUADRATIC SURDS. 

287. A Quadratic Surd is the indicated square root of an 
imperfect square ; as, \J 3, ^ x + 1. 

288. A Binomial Surd is a binomial in which one or both 
of the terms are irrational. 

289. The square root of a rational quantity cannot be 
equal to a rational quantity plus a quadratic surd. 

If possible, let y' a = b + y/ c 

Squaring the equation, a = b 2 + 2 b \j c + c 

or, 2 b \J c = a — b 2 — c 

. a — b 2 — c 
^ C = ^b— 

that is, a surd equal to a rational quantity, which is impossible. 

290. If tin' sum of a rational quantity and a quadratic 
surd be equal to the sum of another rational quantity and 
another quadratic surd, the two rational quantities will be 
equal, also the two quadratic surds. 

That is, if a + \J b — c + \J d 

then a = c and \Jb = \J d 

For, if a is not equal to c, suppose a = c -f- x 
then c-\-x-\-\Jb = c + \Jd 

or, x + s/ b = y/ d 

which is impossible by Art. 289. Hence, a must equal c, and 
consequently y/ b must equal y/ d. 



291. To prove that ifsja + ^b — sj x + \J y, then V ' a — \j b 
= s /x — sfy. 

Squaring the equation \/ a + sjb = \J x + \J y, 
we have tc + \Jb — x + 2 \l~r~y + y 

Whence, by Art. 290, a = x + y and >Jb = 2 \ xy. 



206 ALGEBRA. 

Subtracting these two equations, we have 

a~sJb — x — 2sJxy-\-y 

Extracting the square root, ^ a — ^b = ^ x — ^ y. 

292. To extract the square root of a binomial surd whose 
first term is rational. 

For example, to extract the square root of a + \J b. 

Assume \J a + \fb = \/ x + sj y (1) 

then (Art. 291), ^a—sjb = ^ x — \Jy (2) 

Multiplying (1) by (2), \ja"-b = x-y (3) 

Squaring (1), a + \/b — x + 2^xy + y 

Whence (Art. 290), a = x + y. (4) 

Adding (3) and (4), a + \J~aT^b = 2x, or x= l + ^f~ b - 

Subtracting (3) from (4), a — SJ a 2 — b=2y, or ?/= ^- . 

Substituting these values of x and y in (1) and (2), 

^i^=^(l±^EI) + v /(»=^E»). (5) 

EXAMPLES. 

1. Find the square root of 3 + 2 ^ 2 or 3 + y/ 8. 
Here a = 3 and b = 8. Substituting in (5), we have 



V 3T^=v/( 3 -±^) + V /('^^) 

= v /(^) +v /^H 2 + i >-<- 



RADICALS. 207 

2. Find the square root of 6 — \J 20. 

Here a = 6 and b = 20. Substituting in (6), we have 



293. Examples of this kind may always he solved hy the 
following method : 

3. Extract the square root of 14 — 4 y/ 6. 



y , 14_4 v /6 = V / l^-2v /24 = V / 12-2v / 24 + 2 

= (Art. 116)^12-^2 = 2^3-^2, Arts. 

4. Extract the square root of 43 + 15 \J 8. 



V43 + 15 si 8 = \M3 + sj 1800 = ^43 + 2 sj 450 

= ^25 + 2^450 + 18 = v /25 + V /18 = :5 + 3 v/ 2 > ^ s - 

EULE. 

Reduce the surd term so that its coefficient may he 2. Sep- 
arate the rational term into tiro parts whose product shall be 
the quantity under the radical sign (see first note on page 48), 
writing one part he/ore the surd term and the other part after 
it. Extract the square roots of these parts, and connect them 
by the sign of the surd term. 

The advantage of this method is that it does not require the 
memorizing of formulae (5) and (6). 

EXAMPLES. 

Extract the square roots of the following : 

5. 12 + 2^35. 8. 35 + 10 <J 10. 11. 20-5^12. 

6. 24-2^63. 9. 12 -sj 108. 12. 14 + 3^20. 

7. 16 + 6^7. 10. 8-v/60. 13. 67 -7 si 12. 



208 ALGEBRA. 

Extract the square roots of the following, using formula? 
(5) and (6), Art. 292: 



14. l-12\/-2. 15. 7 + 30V-2. 16. 35-3V/-16. 



17. 2m-2\Jm--n\ 18. x 2 + a x -2 \/ax 3 . 

Extract the fourth roots of the following : 
19. 193 + 22 y/ 72. 20. 17-12)/ 2. 21. 97-56^3. 

SOLUTION OF EQUATIONS CONTAINING RADICALS. 

CASE I. 

294. Wlien there is only one radical term in the equation. 



1. Solve the equation v/ar 2 — 5 — x = — 1. 

Transposing, ^ x 2 — 5 = x — 1 

Squaring, x 2 — 5 = x 2 — 2 cc + 1 

Whence, x = 3, Ans. 

CASE II. 
295. When there are two radical terms in the equation. 



2. Solve the equation \] x — \ f x — 3=1. 
Transposing, \J x — 1 = ^ x — 3 
Squaring, x — 2\Jx + l = x — 3 
Transposing and uniting, — 2 y/ x = — 4 

or, \jx = 2 

Whence, x = 4, Ans. 

CASE III. 
296. When there are three radical terms in the equation. 

3. Solve the equation \J x + 6 + \J x + 13 — v/4a; + 37 = 0. 



RADICALS. 209 

Transposing, \J x + 6 + \/x + l§ = \/4:X + 3T 

Squaring, x + 6 + 2 \/cc 2 + 19.r + T8 + x + 13 = 4 x + 37 



Transposing and uniting, 2 y as 2 + 19 a; + 78 = 2 a; + 18 



or, ^+19 a; + 78 = a: + 9 

Squaring, a; 2 + 19 x + 78 = a;' 2 + 18 a; + 81 

Whence, x = 3, ^f «s. 

RULE. 

297. Transpose the terms of the given equation so that a 
radical term may stand alone in one member ; then raise each 
member to a power of the same degree as the radical. 

If there is still a radical term remaining, repeat the op- 
eration. 

The equation should he simplified as much as possible hefore 
performing the involution. 

Note. All the examples in tliis chapter reduce to simple equations ; 
radical equations, however, may reduce to equations of the second degree, 
for the solution of which see Chapter XXIV. 

EXAMPLES. 

Solve the following equations : 

3/ 



4. V«-8 = 3. 6. y'3a; + 4 + 3 = 6. 8. 8-2^ 



x 



5. V^-3 = 2. 7. ^3-1-2 = 1. 9. o~\2x = 3. 
10. y/4ar-19-2a- = -l. 14. 6 + ^x = \Jl2 + 






11. ^^-3^+6-1=1-^. 15. V / -'-32 + v /a-r=16. 



12. yx* — 6x 2 + 2 = x. 16. ^x — 3— Va; + 12=— 3. 



13. ^ + ^ + 5 = 5. 17. \l2x-l+\J2x + 9 = 8. 



18. ^3x + 10-^3x + 25 = -3. 



19. ^x 2 -3x + 5-\'x 2 -5x-2 = l. 



210 ALGEBEA. 

20. \Jx 2 + 4 x + 12 + \jx 2 - 12 x - 20 = 8. 



21. v^— ^-3 = — . 

y x 

22. ^3^+^3^+13 = ^ ==. 

V o x + 13 

23 V 7 ^ — 3 _y/g; — 4 
V /tc + T _ "y'ic + 1' 

24 y/^_+38_ y/a; + 28 

\/# + 6 \J x + 4 ' 



25. ^-1 + ^ + 4=^4^ + 5. 



26. yWl + V*-2-V / 4jc-3 = 0. 



27. \j2x-3- ^8^ + 1 + VlS ^-92 = 0. 



28. y^ - 3 - V^ - 14 - y/4 a - 155 = 0. 

29. x- \^ (9 + x \/^~~3) = 3. 

30. x + l = \f(l + x \/^ r +T6). 

31 v5^y/3_v^+_3 
V / 2^"- v /2~V / ^ + 2' 

32. y( a *-3a>x + x 2 \/3^~x) = a-x. 



XXIV. — QUADRATIC EQUATIONS. 

298. A Quadratic Equation, or an equation of the second 

degree (Art. 164), is one in which the square is the highest 
power of the unknown quantity ; as, 

ax 2 = b, and x* + 8 x = 20. 

299. A Pure Quadratic Equation is one which contains 
only the square of the unknown quantity ; as, 

ax- = b; and a; 2 = 400. 



QUADEATIC EQUATIONS. 211 

Equations of this kind are sometimes called incomplete equa- 
tions of the second degree. 

300. An Affected Quadratic Equation is one which con- 
tains both the square and first power of the unknown quan- 
tity ; as, 

x 2 + 8 x = 20 ; and a x 2 + b x — e = b x 2 — a x + d. 

Equations of this kind, containing every power of the un- 
known quantity from the first to the highest given, are some- 
times called complete equations. 



PURE QUADRATIC EQUATIONS. 

301. A pure quadratic equation can always he reduced to 
the form 

x 2 = a, 

in which a may represent any quantity, positive or negative, 
integral or fractional. Thus, in the equation 

20a; 2 , K 9 Jv 41 3-5x 2 

Clearing of fractions, 80 x 2 - 12 (5 x 2 + 4) = 41 - (9 - 15 x 2 ) 
or, 80 x 2 - 60 x 2 - 48 = 41 - 9 + 15 x 2 

Transposing and uniting terms, 5 x 2 = 80 

x 2 = 16 

which is in the form x 2 = a. 

Equations of this kind have, therefore, sometimes been de- 
nominated binomial, or those of two terms. 

302. An equation of the form 

x 2 = a 

may he readily solved by taking the square root of each mem- 
ber. Thus, 

x = ± \Ja, 



212 ALGEBKA. 

where the double sign is used, because the square root of a 
quantity may be either positive or negative (Art. 237). 

Note. It may seem at first as though we ought to write the douhle sign 
before the square root of each member, as follows : 

±x = ± y/a. 

We do not omit the double sign before the square root of the first member 
because it is incorrect, but because we obtain no new results by consid- 
ering it. The equation ± x = ± y/ a can be written in four different ways, 

thus, 

x = ^a 

— x=tfa 

-x— - y/« 

where the last two forms are equivalent to the first two, and become iden- 
tical with them on changing all the signs. Hence it is sufficient, in 
extracting the square root of both members of an equation, to place the 
double sign before one member only. 

5x 2 
303. 1. Solve the equation 3 x 2 + 7 = —r- + 35. 

Clearing of fractions, 12 x 2 + 28 = 5 x 2 + 140 

Transposing and uniting terms, 7 x 2 = 112 

x 2 = 16 
Extracting the square root of both members, 

x = ± 4, Ans. 

RULE. 
'Reduce the given equation to the form x 2 = a,and then 
extract the square root of both members. 



EXAMPLES 

Solve the following equations : 
2. 4a; 2 - 7 = 29. 4. 



3. 5x 2 + 5 = 3x 2 +5o. 5. 



t X 1 — 


■5 = 


Sx 


2 


-11 


5 


8 
"3 




5 




4 + x 


4 


— 


X 



QUADRATIC EQUATIONS. 213 

245 



x 



= 5x. 7. 13-V/3a; 2 +lG = 6. 



G 



8. x +^x* + 3= ,-z—z 
y x- + 6 



9. 



_y/3 



l_y/l_aj a 1 + y/l- 



a- 



.'• 



_ cc 2 5 a; 2 _ 7 2 335 

10, Y~ + U^2l~ X + ~2l' 

11. 2 (x - 3) (x + 3) = (x + l) 2 - 2 x. 



12. aa; 2 + 5 = c. 13 



x 2 — b x 2 — a 



AFFECTED QUADRATIC EQUATIONS. 

304. An affected quadratic equation may always be reduced 
to the form 

x 2 -\-jpx = q, 

where p and q represent any quantities, positive or negative, 
integral or fractional. Thus, in the equation 

3x — 3 _ 3x — 6 

5 x 5- = 2 x H s — 

x — J 

Clearing of fractions, 

10 x (x _ 3) _ (6 x - 6) = 4 x (x - 3) + (3 x - 6) (a; - 3) 

or, 10 a- 2 - 30 x - 6 x + 6 = 4 x 2 — 12 a: + 3 a 2 - 15 a; + 18 

Transposing and uniting terms, 3 a; 2 — 9 x = 12 

Dividing by 3, x 2 — 3 x = 4 

which is in the form x 2 + p x =■ q. 

Equations of this kind have, therefore, sometimes been de- 
nominated trinomial, or those of three terms. 



214 ALGEBRA. 

305. Let it be required to solve the equation 

x 2 -\-px = q. 

Equations of this kind are solved by adding to both members 
such a quantity as will make the first member a perfect square, 
and taking the square root of the resulting equation. The pro- 
cess of adding such a quantity to both sides as will make 
the first member a perfect square, is termed Completing the 
Square. 

In any trinomial square (Arts. 104 and 105), the middle 
term is twice the product of the square roots of the extreme 
terms ; therefore the square root of the last term must be 
equal to half the second term divided hy the square root of 
the first. Hence the square root of the quantity which must 

be added to x~ + p x to render it a perfect square, is ~- —- x, 
or — . Adding to both members the square of ^, or ^-, we have 
2 , _,, , P , p _±q+p 



X<+px + - = q + ^ = - 
Extracting the square root of both members, 



X + P = ± \/±<I+P' 






or, x = — 7 -± 



p ^iq+jr 



2 - 1 - 2 



Thus, there are two values of x, 



p \ 4: q + p' 2 ■ p SJ 4 q + p 2 



x = -t>+ —o i or 



2 ' 2 ' 2 2 

We observe from the preceding investigation that the quan- 
tity to be added to complete the square is found by taking half 
the coefficient of cc, and squaring the result. 

Hence, for solving affected quadratic equations, we have the 
following 



QUADKATIC EQUATIONS. 215 

RULE. 

Reduce the equation to the form x 2 + p x = q. 

Complete the square by adding to both members the square 
of half the coefficient of x. Extract the square root of both 
members, and solve the simple equation thus found. 

1. Solve the equation x 2 — 3 x = 4. 

Completing the square, by adding to both members the 

.3 9 

square of -, or -, 

o 9 . 9 25 
x 2 — 3x + - = 4:+ - = — 
4 4 4 

■^ 3 5 

Extracting the square root, x — ~ = ± ^ 

™ 3 5 

Transposing, a; = - ± - 

Taking the upper sign, x = -r+ k = h — 4. 

Z ju Li 

3 5 2 
Taking the lower sign, x= 7> — ^ = — 7 , = — 1. 

Ll Li Li 

Ans. x = £ or — 1. 
We may verify these values as follows : 
Putting x = 4 in the given equation, 16 — 12 = 4. 
Putting x = -l, 1 + 3 = 4. 

These results being identical, the values of x are verified. 

2. Solve the equation 3 x 2 + 8 x = — 4. 

8x4 
Dividing through by 3 x 2 + -— -=— - 

o o 



216 ALGEBRA. 

Completing the square, by adding to both members the square 

,4 16 

0f 3' 0r y 

2 8^ 16 _ 4 16_4 

4 2 

Extracting the square root, x + - = ± - 

o o 

m 4 2 

1 ransposing, x = — - ± - 



rr i • n 4 2 

lakmg the upper sign, x = — - + - = — - 

Taking the lower sign, x = — ^ — ^ = — ^ = — 2. 



2 
.4ns. cc = — - or — 2. 
o 

3. Solve the equation — 3x 2 — 7 x = -=-. 

o 

7 x 10 
Dividing through by — 3, x 1 + -~- = — — 

Completing the square, by adding to both members the square 

. 7 49 
of 6'° r 36' 







X 2 


+ 


7x 49 10 49 __ 9 
X + 36" ~~9~ + 36~36 


Extracting the 


square 


root, 






7_ 3 

X ~t~ „ it A 

b b 


Transposing, 










7 3 
X ~ 6 ± 6 


Whence, 










2 5 A 

x = — q or — k ) -4«& 

«5 O 



A SECOND METHOD OF COMPLETING THE SQUARE. 

306. Although any affected quadratic equation may be 
solved by the method of Art. 305, since its rule is general, 



QUADRATIC EQUATIONS. 217 

still it is sometimes rilore convenient to employ a second 
method of completing the square, known as the " Hindoo 
Method." 

An affected quadratic, reduced to three terms, and cleared 
of all fractions, may he reduced to the form 

a x 2 + b x = c. 
Multiplying each term by 4 a, we have 

By an operation similar to that of Art. 305, we may show 
that b 2 must be added to both members, in order that the first 
member may be a perfect square. Thus, 

4 a 2 x- + 4 a b x + b 2 = b 2 + 4 a c 



Extracting the square root, 2 ax -\- b = ± \ b 2 -\- 4ta c 



Transposing, 2 ax = — b ± \ b 2 + 4 a c 



tv -a- i q -h±\?b 2 +4ac 
Dividing by 2 a, x = ^ . 

It will be observed that the quantity necessary to complete 
the square, is the square of the coefficient of x in the given 
equation. Hence the following 

RULE. 

Reduce the equation to the form a x 2 + b x = r. 

Multiply both members of the equation by four times the 
coefficient of x 2 , and add to each the square of the coefficient 
of x in the given equation. 

Extract the square root of both members, and solve the sim- 
ple equation thus produced. 

Note. The only advantage of this method over the preceding is in 
avoiding fractions in completing the square. 

4. Solve the equation 2 x 2 — 7 x = — S. 



218 ALGEBRA. 

Multiplying both members by four times 2, or 8, 

16 x 2 -5Gx = - 24 

Adding to each member the square of 7, or 49, 

16 x 2 — 56 x + 49 = - 24 + 49 = 25 
Extracting the square root, 4 x — 7 = ± 5 
Transposing, 4cc = 7±5 = 12or2 

Dividing by 4, x = 3 or - , ^4«s. 

307. This method is usually to be preferred in solving 
literal equations. 

5. Solve the equation x 2 + (a — 1) x = a. 
Multiplying both members by four times 1, or 4, 

4 x 2 + 4 (a — 1) x = 4 a 

Adding to each member the square of a — 1, or (a — l) 2 , 

4 x 2 + 4 (a - 1) x + (a - l) 2 = 4 a + (a - l) 2 

= a 2 + 2 a + 1 = (a + l) 2 
Extracting the square root, 

2 a; + (a — 1) = ± (a + 1) 
Transposing, 2 cc = — (a — 1) ± (a + 1) 

Taking the upper sign, 2 x = — (a — 1) + (a + 1) 

r= — « + l + «+l=2 



or, 


x = l. 


Taking the lower sign, 


2 x = -(a-l)-(a + l) 




= — a + 1 — a — 1 = — 2 a 


or, 


x — — a. 



Ans. x = 1 or — a. 

308. In case the coefficient of x in the given equation is 
an even number, the rule may be modified as follows : 



QUADRATIC EQUATIONS. 219 

Multiply both members of the equation by the coefficient of 
x 2 , and add to each the square of half the coefficient of x in 
the given equation. 

6. Solve the equation 7 x 2 + 4 x = 51. 

Multiplying both members by 7, 49 x 2 + 28 x = 357 
Adding to each member the square of 2, or 4, 

49 x 2 + 28 x + 4 = 361 
Extracting the square root, 7 x + 2 = ± 19 

Transposing, 7 a; — — 2 ± 19 = 17 or — 21 

17 

Dividing by 7, ic = — or — 3, ^l»s. 



SOLUTION OF QUADRATIC EQUATIONS BY A 

FORMULA. 

309. In Art. 30G, we showed that if a x 2 + b x = c, then 

-b± \Jb 2 + ±ac ... 

X = 2a * (1) 

We may use this as a formula for the solution of quadratic 
equations as follows : 

7. Solve the equation 3 x 2 + 5 x = 42. 
Here a = 3, b = 5, c = 42 ; substituting these values in (1), 





-5±\/25 + 504 


X — 


6 

-5±v/529_-5±23 



8. Solve the equation 110 « 2 — 21 x = — 1. 
Here a = 110, 6 = — 21, c = — 1 j substituting in (1), 



21 ±1^441 -440 21 ±1 1 1 . 
X = ~ -22CT - = -220- = I6 Or ll'^ 



220 ALGEBRA. 

Note. Particular attention must be paid to the signs of the coefficients 
in substituting. 

9. Solve the equation, — x 2 — 6 x = 8. 
Here a = — 1, b = — 6, c = 8; substituting in (1), 



6 ±^36- 32 6±2 
x = —jr = — = — 4 or — 2, Ans. 



RULE. 

Reduce the equation to the form a x 2 + b x = c. 

The value of x is then a fraction, ivhose numerator is the 
coefficient of x with its sign changed, plus or minus the square 
root of the sum of the square of said coefficient, and four times 
the product of the second member by the coefficient of x' 2 ; and 
whose denominator is twice the coefficient of x 2 . 

310. The following equations may he solved hy either of 
the preceding methods, preference being given to the one best 
adapted to the example considered. Special methods and 
devices may also be employed whenever any advantage can 
thereby be gained. 

EXAMPLES. 

Solve the following equations : 

10. x°- + 2x + 7 = 4:2. 16. 26x + lox 2 = -7. 

11. a; 2 - 9 x- 22 = 0. 17. - 40 + x = 6 x\ 

12. x 2 -Sx = -15. 18. 17x = 2x 2 -6. 

13. z 2 +18* = -65. 19. ^+*=_1. 

t 7 9 f 2 

14. Gx*+7 X -3 = Q. 20. x =--^-. 

£ o o 

3x 2 °2 

15. 13 z - 14 = 3 a;'. 21. ~-^ = x . 

5 o 



QUADRATIC EQUATIONS. 221 

22. ^_-"-_£ = a 24. (as-3)(2as + l)=4. 

6 Jo 

23. ^--^ = -^. 25. (*+5)O-5)-(llx+l)=0. 

26. 4x(18x-l) = (10x-l) 2 . 

27. (3a:-5) 2 -(> + 2) 2 = -5. 

28. (as-l) 2 -(3as + 8) 2 =(2as + 5) 2 . 

29 2 as_ 5 _21 ^_ 3 4 

29< x + 2-~2- d7 " 5-as 7 7* 

x x — 1_3 _ ft cc + 1 a? + 3 _8 

30 - ^TI— "F""2' ^+2 x + 4 _ 3- 

.t 5 — as 15 on 3 a? 2 1 — 8 x x 

31. -3 = -j- . OV. 



5 — x . x 4' ' x — 7 10 5 

5 3z + l 1 . n 2as-l 3x 1 

. = - . 40. 1 — 

x a; 2 4" x 3 a; — 12 



cc 



33. = -— — -. 41. \/20 + x-x 2 = 2(x-5). 

3x+4 4x+l T 



34. _ x __— — - = 0. 42. as+V5as + 10 = 8. 
3a; + 4 7 as- 4 

._ . 35-3as ,. AQ as*-a* + 7 n 

35. 6 as H = 44. 43. = a; + — 

a; ar + 3 a; — 1 o 

14 — r 7 3 22 

36. 4 a -=t-,? = 14 44. -^ ^=^- 

a; + 1 x 2 — 4 a; + 2 o 

45. ^— +- = — - - + 



x* — 1 3 3 (as — 1) a; + 1 



222 ALGEBRA. 

._ a + 3 x — 3 2 a; — 3 

46. -s+- -s = r - 

a + 2 x — 2 x — 1 

,„ 35 + 2 a — 2 2a + 16 

47. T H =-=- -=-, 

a — 1 a + 1 a + 5 

12 + 5 a 2 + a 1 



12 — 5a a 1 — 5a 



49. 


\J X 


+ 1 


— ya- 


-1 


a 


\Jx 

X ■+ 

4/9 


+ 1 

■ V2 


+ v^- 


-1 


"2* 


50. 


(5x4- 

- V^ 


3) = 


= 9. 


K1 


a 2 + 36 a 



a 

52. a ex 2 — b ex + a d x — b d. 

53. a 2 - 2 a a + a 2 - & 2 = 0. 
2 a (a — a) a 



54 



3a-2a 4 



1 111 

55. - = - + - + -. 

a + b + x a b a 

56. (3 a - 2) (a + 5) - (a - 6) (5 a - 16) =301. 

57. (2 a + 3) (3 a + 4) = (8 + a) (2 a + 9). 

58. (2 a - 5) 2 - (2 a - 1) 2 = 8 a - 5 a 2 - 5. 

59. x- + b x + c x = (a + c) (a — b). 
Sa-x 6 a 2 + a b - 2 b 2 b 2 x 



60. abx 2 + 



c 2 



61. (3 a 2 + b 2 ) (a 2 - a + 1) = (3 6 2 + a 2 ) (a 2 + a + 1). 



PROBLEMS. 223 



XXV. — PROBLEMS 

LEADING TO PURE OR AFFECTED QUADRATIC EQUATIONS 
CONTAINING BUT ONE UNKNOWN QUANTITY. 

311. 1. I bought a lot of flour for $ 175 ; and the number 
of dollars per barrel was to the number of barrels, as 4 to 7. 
How many barrels were purchased, and what was the price of 
each ? 

Let x = the number of dollars per barrel, 

7 x 
then - = the number of barrels. 

4 

7 x 2 
By the conditions, —r- = 175 

Whence, x = ± 10. 

Only the positive value is applicable, as the negative value 
does not answer to the conditions of the problem. 

That is, x = 10, the number of dollars per barrel, 

7 x 
and - = 171, the number of barrels. 

4 

2. There is a certain number, whose square increased by 30, 
is equal to 11 times the number itself. Required the number. 

Let x — the number. 

By the conditions, x 2 + 30 = 11 x 
Solving this equation, x = 5 or 6. 

That is, the number is either 5 or G, for each of these values 
satisfies the conditions of the problem. 

3. I bought a watch, winch I sold for $ 56, and thereby 
gained as much per cent as the watch cost me. Required 
the amount paid for it. 

Let x = the amount paid, in dollars. 

Then x = the gain per cent, 

x X" 

and — — X x = — — - = the whole gain in dollars. 



224 ALGEBRA. 



x 2 



By the conditions, -^-r— = 56 — x 

J 100 

Solving this equation, x = 40 or — 140. 

Only the positive value of x is here admissible, as the nega- 
tive result does not answer to the conditions of the problem. 
The cost, therefore, was $ 40. 

Note. When two answers are found to a problem, they should be ex- 
amined to see whether they answer to the conditions of the problem or not. 
Only those which answer to the conditions should be retained. 

PROBLEMS. 

4. I have three square house-lots, of equal size. If I were 
to add 193 square rods to their contents, they would be equiv- 
alent to a square lot whose sides would each measure 25 rods. 
Required the length of each side of the three lots. 

5. There are two square fields, the larger of which contains 
25,600 square rods more than the other, and the ratio of their 
sides is as 5 to 3. Required the contents of each. 

6. Find two numbers whose sum shall be 15, and the sum 
of their squares 117. 

7. A person cut and piled two ranges of wood, whose united 
contents were 26 cords, for 356 dimes ; and the labor on each 
of them cost as many dimes per cord as there were cords in its 
range. Required the number of cords in each range. 

8. A grazier bought a certain number of oxen for $ 240, and 
having lost 3, he sold the remainder at $8 a head more than 
they cost him, and gained $59. How many did he buy ? 

9. The plate of a rectangular looking-glass is 18 inches by 
12, and is to be framed with a frame all parts of which are of 
equal width, and whose area is to be equal to that of the glass. 
Required the width of the frame. 

10. A merchant sold a quantity of flour for §.'59, and gained 
as much per cent as the flour cost him. What was the cost of 
the flour ? 



PROBLEMS. 225 

11. There are two numbers whose difference is 9, and 
whose sum multiplied by the greater is 266. What are the 
numbers ? 

12. A and B gained by trade $ 1800. A's money was in 
the firm 12 months, and he received for his principal and gain 
$2600. B's money, which was $3000, was in the firm 16 
months. What money did A put into the firm ? 

13. A merchant bought a quantity of flour for $ 72, and 
found that if he had bought 6 barrels more for the same 
money, he would have paid $> 1 less for each barrel. How 
many barrel's did he buy, and what was the price of each ? 

14. A square courtyard has a gravel-walk around it. The 
side of the court wants 2 yards of being 6 times the breadth 
of the gravel-walk, and the number of square yards in the walk 
exceeds the number of yards in the perimeter of the court by 
164. Required the area of the court. 

15. My gross income is $ 1000. After deducting a percent- 
age for income tax, and then a percentage, less by one than 
that of the income tax, from the remainder, the income is 
reduced to 8 912. Required the rate per cent at which the 
income tax is charged. 

16. The sum of the squares of two consecutive numbers is 
113. What are the numbers ? 

17. Find three consecutive numbers such that twice the 
product of the first and third is equal to the square of the 
second, increased by 62. 

18. I have a rectangular field of corn which consists of 6250 
hills ; and the number of hills in the length exceeds the num- 
ber in the breadth by 75. How many hills are there in the 
length and breadth ? 

19. A certain company agreed to build a vessel for $ 6300 ; 
but, two of their number having died, those that survived had 
each to advance $ 200 more than they otherwise would have 
done. Of how many persons did the company at first consist ? 



226 ALGEBRA. 

20. A detachment from an army was marching in regular 
column, with C men more in depth than in front; hut Avhen 
the enemy came in sight, the front was increased by 870 men, 
and the whole was thus drawn up in 4 lines. Required the 
number of men. 

21. A has two square gardens, and the side of the one ex- 
ceeds that of the other by 4 rods, while the contents of both 
are 208 square rods. How many square rods does the larger 
garden contain more than the smaller ? 

22. A certain farm is a rectangle, whose length is twice its 
breadth; but should it be enlarged 20 rods in length and 24 
rods in breadth, its contents would be doubled. Of how many 
acres does the farm consist ? 

23. A square courtyard has a rectangular gravel-walk 
around it. The side of the court wants one yard of being six 
times the breadth of the gravel-walk, and the number of square 
yards in the walk exceeds the number of yards in the perim- 
eter of the court by 340. What is the area of the court and 
width of the walk ? 

24. A merchant bought 54 bushels of wheat, and a certain 
quantity of barley. For the former he gave half as many 
dimes per bushel as there were bushels of barley, and for the 
latter 4 dimes per bushel less. He sold the mixture at $ 1 per 
bushel, and lost $ 57.60 by his bargain. Required the quan- 
tity of barley, and its price per bushel. 

25. A lady wishes to purchase a carpet for each of her square 
parlors ; the side of one of them is 1 yard longer than the 
other, and it will require 85 sqxiare yards for both rooms. 
What will it cost the lady to carpet each of the rooms with 
carpeting 40 inches wide, at 81.75 per yard ? 

26. A man has two square lots of unequal dimensions, con- 
taining together 15,025 square feet. If the lots were contigu- 
ous to each other, it would require 530 feet of fence to embrace 
them in a single enclosure of six sides. Required the area of 
each lot. 



QUADRATIC EQUATIONS. 907 

27. A certain number consists of two digits, the left-hand 
digit being twice the right-hand ; and if the digits are inverted, 
the product of the number thus formed, increased by 11, and 
the original number, is 4956. Find the number. 

28. A man travelled 108 miles. If he had gone 3 miles 
more an hour, he would have performed the journey in 6 hours 
less time. How many miles an hour did he go ? 

29. A cistern can be filled by two pipes running together in 
2 hours 55 minutes. The larger pipe by itself will fill it 
sooner than the smaller by 2 hours. What time will each pipe 
separately take to fill it ? 

30. A set out from C towards D, and travelled 3 miles an 
hour. After he had gone 28 miles, B set out from D towards 
C, and went every hour ^ of the entire distance ; and after 
he had travelled as many hours as he went miles in an hour, 
he met A. Required the distance from C to D. 

31. A courier proceeds from P to Q in 14 hours ; a second 
courier starts at the same time from a place 10 miles behind P, 
and arrives at Q at the same time as the first courier. The 
second courier finds that he takes half an hour less than the 
first to accomplish 20 miles. Find the distance from P to Q. 



XXVI. — EQUATIONS IN THE QUADRATIC 

FORM. 

312. An equation is in the quadratic form when it is ex- 
pressed in three terms, two of which contain the unknown 
quantity ; and of these two, one has an exponent twice as 
great as the other. As, 

x 6 — 6 x 3 = 16, 

x s + x* = 72, 
(x- - 1) 2 + 3 {x- - 1) = 18, etc. 



228 ALGEBRA. 

313. The rules already given for the solution of quadratics 

will apply to equations having the same form. For, in the 

equation 

a x 2n + b x n = c, 

let x n = y ; then x 2n = if. Substituting, 

ay 2 +by = c 
Whence, hy Art. 309, we have 



or, x 



• 


— b± \Jb 2 + kac 


U — 


2a 


/*n — 


— b±\Jb 2 + 4tac 




2 a 



from which equation x may he found by extracting the wth 
root of both members. 

314. 1. Solve the equation sc 4 — 5 x 2 = — 4. 

The equation may be solved as in Art. 313, by representing 
x- by y. A better method, however, is the following : 

Completing the square, x* — 5.r 2 + -r- = — 4= + ~r — t 

Extracting the square root, • x 2 — - = ± - 

Transposing, a; 2 = 7 r±- = 4orl 

Whence, x = ± 2 or ±1, Ans. 

2. Solve the equation x 6 — 6 x s = 16. 

Completing the square, x 6 — 6 x a + 9 = 16 + 9 = 25 
Extracting the square root, x a — 3 = ± 5 

Transposing, cc 3 = 3±5 = 8or — 2 

Whence, x = 2 or — \/2, Ans. 



QUADRATIC EQUATIONS. 229 

Here, although the equation is of the sixth degree, we find 
hut two roots. The equation in reality has six roots, hut this 
method fails to give more than two. It will he shown here- 
after how to obtain the other four. 

3. Solve the equation x + 4 \Jx = 21. 
Writing the radical with a fractional exponent, 

x + 4x^ = 21 
which is in the quadratic form. 
Completing the square, cc + 4^:r + 4 = 21 + 4 = 25 

Extracting the square root, \jx + 2 = ± 5 

Transposing, y/x = — 2 ± 5 = 3 or — 7 

Whence, squaring, x = 9 or 49, Arts. 

7 1 

4. Solve the equation 3 x 2 + x^ = 3104 x*. 

Dividing hy x^, 3x% + x% = 3104 

which is in the quadratic form. 

Multiplying hy four times 3, or 12, 

36 x% + 12 x% = 37248 
Completing the square, 36 x% + 12 x% + 1 = 37249 
Extracting the square root, 6 x® + 1 = ± 193 

Transposing, 6 x% = — 1 ± 193 = 192 or - 194 

5 97 

Dividing hy 6, x* = 32 or — y 

\ /97\i 

Extracting the fifth root, x b = 2 or — ( jr-J 

Raising both members to the sixth power, 

x =64 or (-4)°> Ana. 



230 ALGEBRA. 

EXAMPLES. 

Solve the following equations : 

k t 

5. .x 4 + 4cc 2 = 117. 11. 3 a;* -^- = -592. 

6. a-" 4 - 9 or" 2 + 20 = 0. 12. a 3 -a: 2 =56. 

7. a 10 + 31 a- 5 -10 = 22. 13. x-2-s/x = 0. 



8. 81 x- + -4 = 82. 14. x% + x* = 756. 

x- 

9. *• + !??? -14 = 60. IS. ^+2 = i=V» 

a; 2 4 + ya: ya: 



3^* 



2 



10. a; 6 + 20 a; 3 -10 = 59. 16. — — = ^-. 

x — 5 20 



17. Solve the equation (x - 5) 3 - 3 (x - 5) 2 = 40. 

a 9 9 169 

Completing the square, (a; — 5) 3 — 3 (x — 5) 2 + - = 40 + - = — — 

a 3 ^3 
Extracting the square root, (x — 5)* — ■= — ± -=- 

3 3 13 

Transposing, (a; — 5) ^ = - ± — = 8 or — 5 

Squaring both members, (x — 5) 3 = 64 or 25 

Extracting the cube root, x — 5 = 4 or \J 25 

Whence, x = 9 or 5 + \J 25, 



QUADRATIC EQUATIONS. 231 

Solve the following equations : 

18. (x 2 -oa : ) 2 -8(x 2 -5x)=84:. 

19. (2 x - 1) 2 - 2 (2 x - 1) = 15. 

20. (3 x- -2) 2 - ll(3a; 2 -2) + 10 = 0. 

21. (;r J -5) 2 + 29 3 -5)=:96. 

22. Solve the equation x* + 10 x 3 + 17 x 2 — 40 aj — 84 = 0. 
We may write the equation in the form 

x* + 10 x 3 + 2ox 2 — 8x 2 -A0x = 84 

or, (;z 2 +5a;) 2 -8(cr + 5a-)=84 

Completing the square, (x 2 +ox) 2 —8(x 2 +5x) + 16 = 100 
Extracting the square root, (x 2 + 5 x) — 4 = ± 10 

Transposing, (a; 2 + 5 x) = 4 ± 10 = 14 or — 6 

Taking the first value, we have x 2 + 5 x = 14 

Whence (Art. 309), x = ~ 5 ±^ 5 + 56 = =M? = 2 or - 7. 

Taking the second value, we have x 2 + 5 x = — 6 

w, _5±v/25-24 -5±1 

Vv hence, a; = ^ = - = — J or — 3. 

Ans. x = 2, — 7, — 2, or — 3. 

Note. In solving equations of this form, our object is to form a perfect 
trinomial square with the ,r* and x 3 terms, and a portion of the x 2 term. 
By Art. 305, we may effect this by separating the x 2 term into two parts, one 
of which shall be the square of the quotient obtained by dividing the x 3 
term by twice the square root of the x* term. 



232 ALGEBRA. 

Solve the following equations : 

23. x* -12 x 3 + 3±x 2 + 12cc = 35. 

24. x i + 2x 3 -2o x- - 26 x + 120 = 0. 

25. x 4 - 6 cc 3 - 29 x 2 + 114 x = 80. 

26. a 4 + 14 x 3 + 47 x- - 14 a; - 48 = 0. 



27. Solve the equation 2 ar + \'2 x 2 + 1 = 11. 



We may write the equation, (2 x 2 + 1) + y 2 x' 2 + 1 = 12 



49 



Completing the square, (2 x 2 + 1) + V 2 x 2 + 1 + - = -j- 



1 7 

Extracting the square root, V^^+l + T^i^ 



_ 1 7 

Transposing, V 2 ^ + * = ~ 2 ± 2 =r3 ° r_4 

Squaring, 2 a; 2 + 1 = 9 or 16 

Transposing, 2 x 2 = 8 or 15 

15 

Dividing by 2, a; 2 = 4 or ~^- 

/15 
Whence, x = ± 2 or ± t/ — , ^4?is. 

Note. In solving equations of this form, add such quantities to hoth 
members, that the expression without the radical in the first member may 
be the same as that within, or some multiple of it. 

Solve the following equations : 



28. 2 x 2 + 3 x - 5^2^+3 x + 9 = -3. 



29. x- - 6 x + 5 \?x 2 - 6 x + 20 = 46. 



30. 4 z 2 + 6 \/4 ar + 12 x - 2 = - 3 (1 + 4 as). 



31. x 2 - 10 a: - 2 yV 2 - 10 ^ + 18 + 15 = 0. 



32. 3.r 2 +15a:-2v/ar"+5a: + l = 2. 



QUADRATIC EQUATIONS. 233 

XXVII.— SIMULTANEOUS EQUATIONS 

INVOLVING QUADRATICS. 

315. The most general form of an equation of the second 
degree containing two unknown quantities, is 

ax 2 +bxy+cy 2 + dx + ey + f= 0, 

where the coefficients a, b, c, etc. represent any quantities, 
positive or negative, integral or fractional. 

316. Two equations of the second degree containing two 
unknown quantities will generally produce, hy elimination, an 
equation of the fourth degree containing one unknown quantity. 
Thus, if the equations are 

x 2 + y = a 
x + y 2 = b 

From the first, hy transposition, y = a — x 2 ; substituting in 
the second, 

x + (a — x 2 ) 2 = b 

or, x* — 2 a x 2 + x + a 2 — b = 

an equation of the fourth degree. The rules for quadratics 
are, therefore, not sufficient to solve all simultaneous equations 
of the second degree. 

In several cases, however, their solution may he effected hy 
means of the ordinary rules. 

CASE I. 

317. WJien each equation is of the form a x 2 + b y 2 = c. 

1. Solve the equations, 

3x 2 + Aij 2 = 7Q 
3 y 2 - 11 x 2 ^ 4 



234 ALGEBRA. 

Multiplying the first equation by 3, and the second by 4, 

9 x 2 + 12 if = 228 
12 f-Ux 2 = 16 

Subtracting, 53 x 2 = 212 

x 2 = ±, x = ±2. 

Substituting these values in either given equation, 
When x = 2, y = ± 4. 

When a? = — 2, y = ±-4. 

^4«s. # = 2, ?/ = ± 4 ; or, x = — 2, y = ± 4. 

EXAMPLES. 

Solve the following equations : 

2. 2 a; 2 + y 2 = 9 ; 5 x 2 + 6 y 2 = 26. 

3. 4 a; 2 - 3 f = - 11 ; 11 x 2 + 5 y 2 = 301. 

4. 9 x 2 + 24 v/ 2 = 7 ; 72 a; 2 - 180 ?f — - 37. 

5. 20 x a - 16 y 2 = 179 ; 5 x 2 - 336 ?/ 2 = 24. 

CASE II. 

318. When one equation is of the first degree. 

1. Solve the equations, 

x 2 + y 2 = 13 
x + y — 1 

From the second, by transposition, y = 1 — x (1) 

Substituting in the first, x 2 + 1 — 2 x + x 2 = 13 

or, cc 2 — x =6 

AVI / A * o no \ • 1 ± S/T+2l 1 ± 5 _ 

\\ hence (Art. 309), cc = L - = — - — = 3 or — 2. 

Substituting these values in (1), 

When a = 3, y = l — 3 — — 2. 

W T hen x = -2, y = l + 2 = 3. 

Ans. x = 3, y = — 2 ; or, x = — 2, y = 3. 



QUADKATIC EQUATIONS. 235 

In solving examines under Case II, we find an expression 
for the value of one of the unknown quantities in terms of the 
other from the simple equation, which we substitute for that 
quantity in the other equation, thus producing a quadratic 
containing only one unknown quantity, by means of which 
the values of the unknown quantities are readily obtained. 

Note. Although some examples, in which one equation is of the first 
degree (Ex. 1 for instance), may be solved by the methods of the next case, 
yet the method of Case II will be found in general the simplest. 

EXAMPLES. 

Solve the following equations : 

2. x-\r y= — 1; xy = — 56. 
. 3. x + y = 3 ; x 2 + y 2 = 29. 

4. x s — y 3 = — 37 ; x — y = — 1. 

5. x — y = -s--; xy = 20. 

6. 10cc + y = 3ccy;3/ — cc = 2. 

7. x — y = 5; xy = — 6. 

8. x 3 + y 3 = 9 ; x + y = 3. 

9. 3x 2 -2icy = 15; 2« + 3y = 12. 

10. x - y = 3 ; x 2 + y" = 117. 

11. x + y — 11; xy = 18. 

12. x — y = 6; x 2 + y* = 90. 

13. x 3 + y 3 = 152 ; x + y = 2. 

14. x 2 + 3 x y — xf = 23 ; x + 2 y = 7. 

15. x 3 -y 3 = 9S; x-y = 2. 

16. x + y = -±; x 2 + y 2 = 58. 

CASE III. 

319. When the given equations are symmetrical icith 
respect to x and y. 



236 ALGEBRA. 

1. Solve the equations, 



x" + y 2 = 68 
x y = 16 



Multiplying the second by 2, 2 x y = 32 

Adding this to the first equation, x 2 + 2 x y + y' 2 = 100 (1) 
Subtracting it from the first equation, 

x 2 -2xy + y 2 =36 (2) 

Extracting the square root of (1), x + y = ± 10 (3) 

Extracting the square root of (2), x — y — ± 6 (4) 

Equations (3) and (4) furnish four pairs of simple equations, 
x + y = 10 x + y — 10 x + y = — 10 x + y = — 10 
x—y—6 x—y=—Q x—y=6 x—y=—6 



2x = 16 


2x = 4, 


2x = -A 


2x = -16 


x = 8. 


x = 2. 


x = — 2. 


x = — 8. 


y = 2. 


y = 8. 


y = -S. 


y = -2. 



Ans. x = 8, y = 2; x — 2,y = S; 

x = — 2, y = — 8 ; or, x = — 8, y = — 2. 

2. Solve the equations, 

a;3 + ^ = 133 

x 2 — x y + y 2 = 19 

Dividing the first equation by the second, 

x + y = 7 (1) 

Squaring (1), x 2 + 2 x y + y 2 = 49 (2) 

Subtracting the second given equation from (2), 

3 x y = 30 ; or, 4 x y = 40 (3) 

Subtracting (3) from (2), x 2 - 2 x y + y 2 = 9 

Whence, x — y=±3 (4) 



QUADRATIC EQUATIONS. 237 

Adding (1) and (4), 2 x = 10 or 4 

Whence, x = 5 or 2. 

Substituting these values in (1), 
When x = 5, y = 2 

cc = 2, y = 5. 
Ans. x — 5, y = 2 ; or, a; = 2, ?/ = 5. 

The example might have been solved by substituting the 
value of y derived from (1) in either of the given equations, 
as in Case II. 

The student will notice the difference between Examples 1 
and 2 as regards the arrangement of the last portion of the 
work. 

3. Solve the equations, 

x + y = 20 



Multiplying the first equation by 2, 2 x 1 + 2 y 2 = 416 (1) 

Squaring the second equation, x 2 + 2 x y + y 2 = 400 (2) 

Subtracting (2) from (1), x 2 - 2 x y + y 2 = 16 

Whence, x — y = ± 4 (3) 

Adding the second given equation and (3), 

2 x = 24 or 16 
Whence, x = 12 or 8. 

Substituting these values in (3), 

When » = 12, y = 8 

x = 8, y = 12. 
J??5. aj .= 12, y = 8 ; or, a; = 8, y = 12. 

This example is solved more readily by the method of Case 
II; we solve it by Case III merely to show how equations 
may be solved symmetrically, when one is of the first degree. 



238 ALGEBRA. 

EXAMPLES. 

Solve the following equations : 

4. x' 2 + if = 25- xy = 12. 

5. x* + y 2 = S5; xy = 42. 

6. x 3 + y 3 = -19; x 2 -xy + y 2 = 19. 

7. x 3 - y 3 = - 65 ; x 2 + x y + y 2 = 13. 

8. o; + 2/ = l;a;y = — 6. 

9. cc 2 + y 2 = 65 ; a; — y = 11. 

10. a? 9 +y 2 = 61; x + y = ll. 

11. a; 8 — y 8 = 117; a; — y = 3. 

Note. Exs. 8, 9, 10, and 11 are to be solved like Ex. 3, and not by 
the method of Case II. In solving Ex. 11, begin by dividing the first 
equation by the second. 

CASE IV. 

320. When the equations are of the second degree, and 
homogeneous. 

Note. Some examples, in which both equations are of the second de- 
gree and homogeneous, are solved more easily by the methods of Cases I 
and III, than by that of Case IV. The method of Case IV is to be used 
only when the example can be solved in no other way. 

1. Solve the equations, 

x 2 — xy = 35 
xy + y 2 = 18 

Letting y = vx, we have 

35 

x 2 — v x 2 = 35, or x 2 (1 — v) = 35 ; whence, x 2 = — - - — (1) 

18 

v x 2 + v 2 x 2 = 18, or x 2 (v + v 2 ) = 18 ; whence, x 2 = - '■ — r 

v + v 



QUADRATIC EQUATIONS. 239 

Equating the values of x 2 , -z = -= 

10 1 — vv+v 2 

Clearing of fractions, 35 v + 35 v 2 =18 — 18 v 

Transposing and uniting, 35 v 2 + 53 v = 18 

Whence (Art. 309), 



- 53 ± V -'809 + 2520 - 53 ± 73 2 9 

V = 70 = 70 =7° r -5 

2 
If v = - , substituting in (1), x 2 = 49, or x = ± 7 

Substituting in the equation y = v x, 

o 
When x = 7, y=-xT=2 

x = -7, !/ = ~x-7 = -2. 
If v = — ■=, substituting in (1), x 2 = - 7 r- , ov x = ± —j-^ 

£> 2 y 2 

Substituting in the equation y = vx, 

«n 5 9 5 9 

men x =V2' y= ~5 x 72 = ~72 

5 9 5 9 



^2'^ 5' N V 2 V 2 

.4ns. cc = 7, ?/ = 2; £ = — 7, y = — 2; 

5_ 9 j5_ _9_ 

X ~^/2 ,y ~ S/2 ] ° l,X ~ \J2 ,y ~s/2' 

Note. In using the equation y~v x, to calculate the value of y when 
x has been found, care should be taken to use that value of v which n/i* 
used in getting the particular value of x. 

EXAMPLES. 

Solve the following equations : 

2. x 2 + xy + ±y 2 = 6; 3x 2 +8y 2 = U. 

3. 6x 2 -5xy + 2y 2 = 12; 3 x 2 + 2 x y- 3 y 2 = -3. 



240 ALGEBRA. 

4. x 2 + x y = 12 ; x y — y 2 = 2. 

5. 2 y 2 - 4 x y + 3 x 2 = 17 ; y 2 -x 2 = 16. 

6. x 2 + x y — y 2 = 1 ; x 2 — x y + 2 y 2 = 8. 

7. 2 x 2 — 2 a; y — ?/ 2 = 3; x 2 + 3a;y + 2/ 2 = 11. 

321. We append a few miscellaneous examples, for the 
solution of which no general rules can be given. Various arti- 
fices are used ; familiarity with which can only be obtained by 
experience. 

1. Solve the equations, 

x z — y s = 19 

x 2 y — x y 2 = 6 
Multiplying the second by 3, 3 x 2 y — 3 x y 2 = 18 (1) 

Subtracting (1) from the first given equation, 

x z — 3 x 2 y + 3 x y 2 — y 3 = 1 
Extracting the cube root, x — y = 1 (2) 

Transposing, x = 1 + y (3) 

Dividing the second given equation by (2), x y = 6 (4) 

Substituting from (3) in (4), y (1 + y) = 6 

or, y 2 + y = 6 



m -l±V/T+24 -1±5 

Vv hence, ?/ — k — i5 — = ^ or — 3. 

Substituting in (3), 

When y = 2, x = 3 

y = — 3 } x = — 2. 

Ans. x = 3, y = 2 ; or, a; = — 2, ?/ = — 3. 



QUADRATIC EQUATIONS. 241 

2. Solve the equations, 

V * 

x + y = 12 

Let x — u + v, and y = u — v. 

Then x + y = 2 u ; whence, 2u — 12, or u = 6. 
From the first given equation, x z + y s = 18 x y 

Substituting x = 6 + v, and y = 6 — v, we have 

(6 + vf + (6 - v) 3 = 18 (6 + v) (6 - 1>) 
Reducing, 432 + 36 v 2 = 648 - 18 v 2 

Whence, 54v 2 = 216 

v 2 = 4, v = ± 2 
Then a5 = 6+v = 6±2 = 8or4. 

Substituting these values in the second given equation, 

When x = 8, y = 4 

a; = 4, ?/ = 8, ^4ws. 

3. Solve the equations, 

x 2 + y" + x + y = 18 

x ?/ = 6 

Adding twice the second equation to the first, 

x 2 + 2 x y + y 2 + x + y — 30 
or, (x + y) 2 +(;x + y)=30 

N - 1 ± Vl + 120 -1±11 r r 

Whence, (x + y) = ^ = „ = 5 or — 6. 

Taking the first value, x + y = 5 (1) 

and the second given equation, xy = 6 (2) 

From (1), y=5—x ; substituting in (2), x 2 — 5 x = — 6 

5±\/25-24 5±1 
Whence, x = —^ = — „ — = 3 or 2. 

Substituting in (1), 

When, x = 3, y — 2 

cc = 2, y = S. 



242 ALGEBRA. 

Taking the second value, x + y = — 6 (3) 

and the second given equation, x y = 6 (4) 

From (3), y — — C — x ; substituting in (4), 

x 2 + 6 x = — 6 

_6±y/36-24 -6+2^3 /0 
Whence, a; = ^_ = =r_VL_ — _ 3 ± ^ 3. 

Substituting in (3), 

When a= — 3 + v/ 3 ; t/ = — S-—^3. 

x = -3-sJ3,y = — 3 + sJ3. 
Ans. x = 3, y = 2 ; cc = 2, ?/ = 3 ; 

a; = — 3+\/3, y=— 3— ^3; or, x=— 3— y/3, ?/=— 3 + y/3 

4. Solve the equations, 

x* + 7/ = 97 (1) 

* +y =~1 (2) 

Raising (2) to the fourth power, 

x i + 4 x s y + 6 x 2 y 2 + 4 x y 3 + y 4 = 1 (3) 
Subtracting (1) from (3), 4 a? 3 y + 6 ar ?/ 2 + 4 a; ?/ 3 = — 96 
or, 3 a; 2 .?/ 2 + 2 a; ?/ (x 2 + ?/ 2 ) = - 48 (4) 

But from (2), x 2 + y 2 = l-2xy 

Substituting in (4), 3 x 2 y 2 + 2 x y (1 - 2 x y) = - 48 
or, a: 2 y 2 — 2 cc y = 48 

Whence, x y = = — — — = — 6 or 8. 

Taking the first value, x y = — 6 

From (2), y = — 1 — x ; substituting, x 2 + x = 6 

Whence, x = j- — = — - = 2 or — 3. 

Substituting in (2), 

When x = 2, y = — 3. 

x = — 3, y = 2. 

Taking the second value, x y = 8 



QUADRATIC EQUATIONS. 243 

From (2), y = — 1 — x ; substituting, x 2 + x = — 8 

-l±y/T^32 -l±y/-3i 

\\ hence, x = -= = „ . 

Substituting in (2), 

- 1 + V /=7 31 - 1 - V^3l 
When x = ^ > V~ <j ' 

- 1 - y/^31 - 1 + \/^31 

*= — 2 — ->y=- —2 — 

Ans. x = 2,y = ~3; x--S,y = 2; x= - 1+ ^~ 31 ^ 

_1_ S /Z3i _i_y/Z3i _i +v /I73i 
y = y — i ov > x = 2~ ->y= 2 ' 

EXAMPLES. 

Solve the following equations : 

5. x + y = 9; \/x+\/y = 3.. 

6. x + \J~x~y + y = 19 ; x 2 + x y + f = 133. 

7. a; 2 y + ar y 2 = 30 ; x* if + x~y i = 468. 

8. x 2 + y 2 — 'x - y = 18 ; x y + x + y = 19. 

9. x 2 + 3x + y = 73 - 2 x y ; y 2 + 3 y + x = 44. 

. « n „ 5 x 1/ xy 

10. as» + I f=_£j x - y = -f. 

♦ J 4 

„, a; 4v/# 33 „ 

11. - + -7— = -r ; » — y=& 

y \jy 4 

12 - 2 + 8 = li 5 + S=- 4 

13. x 2 y + x y 2 = 30 ; a 3 + y s = 35. 

14. cc+V / *'?/=3;?/+V«2/ = — 2 

15. x 2 y + y 2 x = 6; - + - = -. 

16. r 4 + ?/ 4 = 17; x — y = 3. 

17. z 5 - y 5 = - 211 ; cc — y = — 1. 

18. ar + y 2 = 7 + as y ; x s + y s = 6 x y — 1. 

19. 2x 2 -7x?/-2?/' 2 =:5; 3x y-x 2 + 6 y 2 = U. 



244 ALGEBRA. 

20. -A^ + Jl^S | + | = 2. 
y + 3 x + J 2 2 o 

21. z + 2 = 7; 2y-3* = -5; a; 2 + y 1 - z~ = 11. 

22. xz = y 2 ; (x + y)(z—x—y) = 3; (x + y + z) (z— x— y) = 7. 



XXVIII. — PROBLEMS 

LEADING TO SIMULTANEOUS EQUATIONS INVOLVING 

QUADRATICS. 

322. 1. What two quantities are those, the sum of whose 
squares is 130, and the difference of whose squares is 32 ? 

Let x = one number, 

and y = the other. 

By the conditions, x 2 + y 2 = 130 

x 2 -y 2 = 32 
Solving these equations, as in Case I, Art. 317, 

x = 9, y = ± 7 ; # 

or, x = — 9, y = ± 7. 

This indicates four answers to the problem : 

9 and 7, 
9 and — 7, 

— 9 and 7, 

— 9 and — 7. 

Any one of these pairs of values will satisfy the conditions 
of the problem. 

2. A says to B, " The sum of our money is $ 18." B re- 
plies, "If twice the number of your dollars were multiplied by 



PROBLEMS. 245 

mine, the product would be $ 154." How many dollars had 
each ? 

Let x = A's dollars, 

and y = B's. 

By the conditions, x + y = 18 

2 x y = 154 

Solving these equations, as in Case II, Art. 318, 

x = li y = U; 

or, x = 11, y = 7. 

That is, either Alias $7, and B $11, or A has $11, and 

B $7. 

3. The price of two coats and one vest is $ 38. And the 
price of a coat less that of a vest, is to $ 23, as $ 7 is to the 
sum of the prices of a coat and vest. What is the price of a 
coat, and what of a vest ? 

Let x = the price of a coat in dollars, 

and y = the price of a vest. 

By the conditions, 2 x + y = 38 

and x — y : 23 = 7 : x + y 

or (Art. 181), a 2 -?/ 2 = 161 

Solving these equations, as in Case II, Art. 318, 

x = 15, y = 8 ; 

107 100 

x = — ,y = - — . 

Only the first answer is admissible, as a negative value of 
either unknown quantity does not answer to the conditions 
of the problem. Hence, the price of a coat is $15, and of a 
vest, $ 8. 

Note. The note after Ex. 3, Art. 311, applies with equal force to the 
problems in this chapter. 



246 ALGEBRA. 

PROBLEMS. 

4. The difference of two quantities is 5, and the sum of 
their squares is 193. What are the quantities ? 

5. There are two quantities whose product is 77, and the 
difference of whose squares is to the square of their difference 
as 9 to 2. Required the quantities. 

6. A and B have each a field, in the shape of an exact 
square, and it requires 200 rods of fence to enclose hoth. The 
contents of these fields are 1300 square rods. What is the 
value of each at $ 2.25 per square rod ? 

7. Two gentlemen, A and B, were speaking of their ages. 
A said that the product of their ages was 750. B replied, that 
if his age were increased 7 years, and A's were diminished 2 
years, their product would be 851. Required their ages. 

8. A certain garden is a rectangle, and contains 15,000 
square yards, exclusive of a walk, 7 yards wide, which sur- 
rounds it, and contains 3696 square yards. Required the 
length and breadth of the garden. 

9. What k two numbers are those whose difference multi- 
plied by the less produces 42, and by their sum, 133 ? 

10. A and B lay out money on speculation. The amount 
of A's stock and gain is $ 27, and he gains as much per cent 
on his stock as B lays out. B's gain is $ 32 ; and it appears 
that A gains twice as much per cent as B. Required the capi- 
tal of each. 

11. I bought sugar at such a rate, that the price of a pound 
was to the number of pounds as 4 to 5. If the cost of the 
whole had been 45 cents more, the number of pounds would 
have been to the price of a pound as 4 to 5. How many pounds 
were bought, and what was the price per pound ? 

12. A and B engage in speculation. A disposes of his share 
for $ 11, and gains as many per cent as B invested dollars. 



PROBLEMS. 247 

B's gain was $ 36, and the gain upon A's investment was 4 
times as many per cent as upon B's. How much did each 
invest ? 

13. A man bought 10 ducks and 12 turkeys for $ 22.50. He 
bought 4 more ducks for $ 6, than turkeys for $ 5. What was 
the price of each ? 

14. A man purchased a farm in the form of a rectangle, 
whose length was 4 times its breadth. It cost £ as many dol- 
lars per acre as the field was rods in length, and the number 
of dollars paid for the farm was 4 times the number of rods 
round it. Required the price of the farm, and its length and 
breadth. 

15. I have two cubic blocks of marble, whose united lengths 
are 20 inches, and contents 2240 cubic inches. Required the 
surface of each. 

16. A's and B's shares in a speculation altogether amount 
to $ 500. They sell out at par, A at the end of 2 years, B of 
8, and each receives in capital and profits $ 297. How much 
did each embark ? 

17. A person has $ 1300, which he divides into two portions, 
and loans at different rates of interest, so that the two por- 
tions produce equal returns. Jf the first portion had been 
loaned at the second rate of interest, it would have produced 
$ 36 ; and if the second portion had been loaned at the first 
rate of interest, it would have produced $49. Required the 
rates of interest. 

18. Two men, A and B, bought a farm of 104 acres, for 
which they paid $320 each. On dividing the land, A says to 
B, "If you will let me have my portion in the situation which 
I shall choose, you shall have so much more land than I, that 
mine shall cost $ 3 per acre more than yours." B accepted the 
proposal. How much land did each have, and what was the 
price of each per acre ? 



248 ALGEBRA. 

19. A and B start at the same time from two distant towns. 
At the end of 7 daj's, A is nearer to the half-way house than 
B is, by 5 miles more than A's day's journey. At the end of 
10 days they have passed the half-way house, and are distant 
from each other 100 miles. Now it will take B 3 days longer 
to perform the whole journey than it will A. Required the 
distance of the towns, and the rate of walking of A and B. 

20. Divide the number 4 into two such parts that the prod- 
uct of their squares shall be 9. 

21. The fore-wheel of a carriage makes 15 revolutions more 
than the hind-wheel in going 180 yards ; but if the circumfer- 
ence of each wheel were increased by 3 feet, the fore-wheel 
would only make 9 revolutions more than the hind-wheel in 
going the same distance. Find the circumference of each 
wheel. 

22. A ladder, whose foot rests in a given position, just 
reaches a window on one side of a street, and when turned 
about its foot, just reaches a window on the other side. If 
the two positions of the ladder are at right angles to each 
other, and the heights of the windows are 36 and 27 feet re- 
spectively, find the width of the street and the length of the 
ladder. 

23. A and B engaged to reap a field for 90 shillings. A 
could reap it in 9 days, and they promised to complete it in 5 
days. They found, however, that they were obliged to call in 
C, an inferior workman, to assist them the last two days, in 
consequence of which B received 3.?. 9d. less than he other- 
wise would have done. In what time could B and C each reap 
the field? 

24. Cloth, being wetted, shrinks J in its length and ,\ ; in its- 
width. If the surface of a piece of cloth is diminished by 5^ 
square yards, and the length of the four sides by 4} yards, 
what was the length and width of the doth originally? 



THEORY OF QUADRATIC EQUATIONS. 249 



XXIX. — THEORY OF QUADRATIC EQUA- 
TIONS. 

323. A quadratic equation cannot have more than two 
roots. 

We have seen (Art. 304) that every complete quadratic 
equation can be reduced to the form 

x 2 + p x = q. 

Suppose, if possible, that a quadratic equation can have three 
roots, and that r 1; r 2 , and r 3 are the roots of the equation 
x°- + p x — q. Then (Art. 166), 

r 1 2 +pr 1 = q (1) 

r 2 2 +pr 2 — q (2) 

r 3 2 +pr 3 — q (3) 

Subtracting (2) from (1), (?y — r.F) + p (r x — r 2 ) = 

Dividing through by r y — r 2 , which by supposition is not 
zero, as the roots are not equal, 

r \ + r 2 +p = 

Similarly, by subtracting (3) from (1), we have 

n + n + p — o 

Hence, r x + r 2 + p = r x + r s -f p 

or, r 2 = r 3 . 

That is, two of the roots are identical. Therefore, a quad- 
ratic equation cannot have more than two roots. 

DISCUSSION OF THE GENERAL EQUATION. 

324. By Art. 305, the roots of the equation x 2 + p x = q 



are 



-p-\- \/p 2 + ±q 1 —p — \Jp 2 +4:q 



250 ALGEBRA. 

1. Suppose q positive. 

Since p 2 is essentially positive (Art. 227), the quantity 
under the radical sign is positive and greater than ,p° ; so 
that the value of the radical is greater than P- Hence, one 
root is positive, and the other negatn r e. 

If p is positive, the negative root is numerically the larger; 
if p is zero, the roots are numerically equal ; and if p is nega- 
tive, the positive root is numerically the larger. 

2. Suppose q equal to zero. 

The quantity under the radical sign is now equal to p 2 ; so 
that the value of the radical is p. Hence, one of the roots is 
equal to 0. The other root is positive when p is negative, and 
negative when p is positive. 

3. Suppose q negative, and 4 q < p 2 . 

The quantity under the radical sign is now positive and less 
than p> 2 ; so that the value of the radical is less than p. 

If p is positive, hoth roots are negative ; and if p> is nega- 
tive, both roots are positive. 

4. Suppose q negative, and 4 q =p 2 . 

The quantity under the radical sign is now equal to zero ; 
so that the two roots are equal ; being positive if p is negative, 
and negative if p is positive. 

5. Suppose q negative, and 4 q > p 2 . 

The quantity under the radical sign is now negative ; hence, 
by Art. 282, both roots are imaginary. 

325. All these cases may be readily verified by examples. 

Thus, in the equation x 2 — 3;r = 70, as p is negative and q 
positive, we should expect to find one root positive and the 
other negative, and the positive root numerically the larger 

And this is actually the rase, for on solving the equation, wfc 
find x = 10 or - 7. 



THEORY OF QUADRATIC EQUATIONS. 251 

326. From the quadratic equation x 2 +px = q, denoting 
the roots by r x and r. 2 , we have 



-p+S/Y + lq _ 1m -p-Sjp' + lq 

2 



n = — ^ , and r 2 



Adding these together, we have 

2p 
r x + r 2 = — -^- = —p. 

Multiplying them together, we have 

n r 2 = £=M±±1± (Art. 106) = - *± = - q . 

That is, if a quadratic equation be reduced to the form 
x 2 + p x = q, the algebraic sum of the roots is equal to the co- 
efficient of the second term, with its sign changed • and the 
product of the roots is equal to the second member, with its 
sign changed. 

327. The equation a x 2 + b x + c = 0, by transposing c, and 
dividing each term by a, becomes 



x 2 -\ 



bx c 

a a 



Denoting the roots of the equation by x l and x 2 , we have, by 
the previous article, 

b _c 

Xi -j- Xo — , and x± x.y — — • 

a a 



328. A Quadratic Expression is a trinomial expression of 
the form a x 2 + b x + c. The principles of the preceding 
article enable us to resolve any quadratic expression into two 
binomial factors. 

The expression a x' 2 + b x + c may be written 

f bx c 

a I x -\ 1 — 

V a a 



252 ALGEBRA. 

b c 

By the previous article, - = —(% + x 2 ), and - =x 1 x 2 , where 

x x and x 2 are the roots of the equation ax 2j rbx + c = Q; 
which, we ohserve, may he obtained by placing the given 
expression equal to 0. Hence, 

ax 2 + bx + c = a [x 2 — (x x + x 2 ) x + x x x 2 ~\. 

The expression in the bracket may be written 

/V»** rtn /yt ry* /y> I /y» sy 

which, by Case II, Chap. VIII, is equal to (x — x^ (x — x 2 ). 
Therefore, a x 2 + b x + c = a (x — x x ) (x — x 2 ). 

1. Factor 6 x 2 + 11 x -+- 3. 

Placing the expression equal to 0, and solving the equation 
thus formed, we find 



_ - 11 + ^121 - 72 _ - 11 ± 7 _ 3 1 

X ~~ ~V1~ ~ _ ~~12 - ~2' 0r ~3- 

Then, a = 6, x 1 = — ^, x 2 = — 5. 

Therefore, 6 x 2 + 11 x + 3 = 6 (x + |) (» + 5) 

= (2 a; + 3) (3 x + 1), Ans. 

2. Factor 4 + 13 x - 12 x 2 . 

Placing the expression equal to 0, and solving the equation 
formed, we have 



_ - 13 ± y/ 169 + 192 _ - 13 ± 19 _ 4 1 

X ~ -24 -24 ~3' ° r 4 

4 1 

Then, a = — 12,x 1 = 7 r, x 2 = — -. 



THEORY OF QUADRATIC EQUATIONS. 253 

Therefore, 4 + 13 x - 12 x 2 = - 12 he - ^J (a; + ^J 

= -3{x-l)4(x + \) 

— (4 — 3 x) (4 a; + 1), ^4?is. 

Note. It should be remembered, in using the formula a (x-x{) (x - a'2 N , 
that a represents the coefficient of x 2 in the given expression J hence, in 
Example 2, we made a= - 12. 

EXAMPLES. 

Factor the following expressions : 

3. x 1 + 73 x + 780. 9. 8z 2 + 18z-5. 

4. x 2 - 11 x + 18. 10. 4 z 2 - 15 * + 9. 

5. x 2 -4cc-60. 11§ 2x 2 +x-6. 

6. a 8 + 10 a; -39. 12. 9x 2 -12a- + l. 

7. 2 a; 2 -7 a; -15. 13. l-8x-x 2 . 

8. 21 a- 2 + 58 a; + 21. 14. 49 x 2 + 14 a- - 19. 

329. The principles of Art. 328 furnish a method of form- 
ing a quadratic equation which shall have any required roots. 

For, the equation a x 2 + b x + c = 0, if its roots be denoted 
by x x and x 2 , may be written, by Art. 328, 



a 



(x — Xy) (x — x 2 ) = 0, or (x — £Cj) (x — a- 2 ) = 9. 



Hence, to form an equation whose roots shall be x v and x.,, 
we subtract each of the two roots from x, and place the product 
of the resulting binomials equal to zero. 

7 
1. Required the equation whose roots are 4 and — -r • 

By the rule, (x — 4) (x + -)= 



254 



or, 



ALGEBRA. 










x 2 - 


9x 


-7 = 


= 




4x 2 - 


9x- 


-28 = 


= o, 


Ans. 



Clearing of fractions, 

EXAMPLES. 

Form the equations whose roots are 

17 

2. 1 and — 2. 5. 7 and — 6 \ . 8. — -^- and 0. 

8 4 

3. 4 and 5. 6. — - and - . 9. 1 + sj 5 and 1 — ^5. 

o i 

4. 3 and —■=. 7. — 2 J and — 3^. 10. m + \/ w and m — \Jn. 

o 

330. By Art. 328, the equation a x 2 + b x + c = may be 
written (x — a^) (x — x 2 ) = 0, if acj and x 2 are its roots;- we 
observe that the roots may be obtained by placing the factors 
of the first member separately equal to zero, and solving the 
simple equations thus formed. 

This principle is often useful in solving equations. 

1. Solve the equation (2x — 3) (3 x + 5) = 0. 

3 

Placing the first factor equal to zero, 2x — 3 = 0, or a; = - . 

A 

5 

Placing the second factor equal to zero, ox+ 5=0, orx= — - . 

A 3 5 

Ans. x = - or — ^ . 

A o 

2. Solve the equation x' 2 + 5 x = 0. 
The equation may be written x (x + 5) = 

Placing the first factor equal to zero, x = 0. 

Placing the second factor equal to zero, x + 5 = 0, or x = — 5. 

Ans. x = or — 5. 



THEORY OF QUADRATIC EQUATIONS. 255 

EXAMPLES. 

Solve the following equations : 

3. (as-|) (as-2) = 0. 9. 2as 8 -18as = 0. 

4. (as+.5)(as-l) = 0. 10. (2as + 5)(3as-l) = 0. 

5. (as-?) (as + ?)=0. 11. (aas + b) (cx-d) =0. 

6. (as + 8)(as + i)=0. 12. (x 2 -4) (as 2 -9) = 0. 

7. 2as 2 -13as = 0. 13. (3 a; + 1) (4 x* - 25) = 0. 

8. 3 as 3 + 12 a: 2 = 0. 14. (as 2 -a)(as 2 -aas-£)=0. 

15. as (2 as + 5) (3 x -7) (4 x + 1) = 0. 

16. (x 2 -5x + 6)(x 2 + 7x+ 12) (2 x* + 9x-5) = 0. 

331. Many expressions may be factored by the artifice of 
completing the square, used in connection with the method of 
Case IV, Chapter VIII. 

1. Factor x i + a\ 

x* + a 4 = x* + 2 x 2 a 2 +a*-2 x 2 a 2 
= (x 2 + a 2 ) 2 - (a x ^/ 2) 2 
= (Art. 117) (x 2 + a x ^ 2 + a 2 ) (x 2 — a x \J 2 + a 2 ), Ans. 

2. Factor x 2 — ax + a 2 . 

as 2 — ax + a 2 = x 2 + 2ax + a 2 — 3ax. 

= (x + a) 2 -(^3axY 



= (x + V^3 a x + a) (x — \J3 a x + a), Ans. 



256 ALGEBRA. 

EXAMPLES. 

Factor the following expressions : 

3. x 2 + l. 5. a 2 -3ab + b*. 7. x 2 -x-l. 

4. x 2 +x+l. 6. x i -lx 2 y 2 -^y\ 8. m i + m 2 n 2 +?i\ 

332. We have seen (Art. 330) that any equation whose 
first member can be factored, and whose second member is 
zero, may be solved by placing the factors separately equal to 
zero and solving the equations thus formed. This method of 
solution is frequently the only one which will give all the roots 
of the equation. 

1. Solve the equation x 3 = 1. 

The equation may be written x 3 — 1 = 0, or (Art. 119), 
(x - 1) (x 2 + x + 1) = 0. 
Placing the first factor equal to zero, 

x — 1 = 0, or x = 1. 
Placing the second factor equal to zero, 

x 2 + x + 1 = 0, or x 2 + x = — 1 



Whence (Art, 309), - - = ^ ± ^ 



x = 



2 2 



Hence, x = 1 or ^1 , Ans. 



EXAMPLES. 

Solve the following equations : 

2. x i = -l. 4. x 4 +« 4 = 0. 6. x 6 = l. 

3. x 3 = -l. 5. X 4- X * + 1 = 0. 7. rr 4 - — +1 = 0. 

These examples afford an illustration of the statement made 
in Art. 167 that the degree of an equation indicates the num- 
ber of its roots. 



DISCUSSION OF PROBLEMS. 257 

XXX. — DISCUSSION OF PROBLEMS 

LEADING TO QUADRATIC EQUATIONS. 

333. In the discussion of problems leading to quadratic 
equations, we find involved the same general principles which 
have been established in connection with simple equations 
(Arts. 205-212), but with certain peculiarities. 

These peculiarities will be now considered. They arise from 
two facts : 

1. That every quadratic equation has two roots • and 

2. That these roots are sometimes imaginary. 

334. In the solution of problems involving quadratics, it 
has been observed that the positive root of the equation is 
usually the true answer ; and that, when both roots are posi- 
tive, there may be two answers, either of which conforms to 
the given conditions. 

The reason why results are sometimes obtained which do 
not apply to the problem under consideration, and are there- 
fore not admissible, is that the algebraic mode of expression 
is more general than ordinary language ; and thus the equa- 
tion which conforms properly to the conditions of the problem 
will also apply to other conditions. 

1. Find a number such that twice its square added to three 
times the number may be 65. 

Let x = the number. 

Then 2 x 2 + 3 x = 65 (1) 

13 

Whence, x = 5 or — . 

The positive value alone gives a solution to the problem in 
the sense in which it is proposed. 

To interpret the negative value, we observe that if we 
change x to — x, in equation (1), the term 3 x, only, changes 



258 ALGEBRA. 

its sign, giving as a result the equation 2 x' 2 — 3 x = 65. Solv- 

13 

ing this equation, we shall find x = — or — 5, which values 

only differ from the others in their signs. We therefore may 

13 

consider the negative solution, — , taken independently of 

Li 

its sign, the proper answer to the analogous problem (Art. 

205): 

" Find a number such that twice its square diminished by 
three times the number may be Go." 

2. A farmer bought some sheep for $ 72, and found that if 
he had bought 6 more for the same money, he would have paid 
$ 1 less for each. How many sheep did he buy ? 

Let x = the number of sheep bought. 

72 
Then — = the price paid for one, 



x 

72 
and = the price paid, if 6 more. 

JO \~ O 

72 72 

By the conditions, — = - + 1 

J ' x x + 6 

Whence, x = 18 or — 24. 

Here the negative result is not admissible as a solution of 
the problem in its present form; the number of sheep, there- 
fore, was 18. 

If, in the given problem, "6 more " be changed to "6 fewer" 
and "$1 less" to "$1 more," 24 will be the true answer. 

Hence, we infer that 

A negative result, obtained as one of the answers to a prob- 
lem, is sometimes the answer to another analogous problem, 
formed by attributing to the unknown quantity a quality 
directly opposite to that which has been attributed to it. 



DISCUSSION OF PROBLEMS. 259 

INTERPRETATION OF IMAGINARY RESULTS. 

335. It lias been shown (Art. 324) under what circum- 
stances a quadratic equation will be in form to produce imagi- 
nary roots. It is now proposed to interpret such results. 

Let it be required to divide 10 into two such parts that their 
product shall be 26. 

Let x = one of the parts. 

Then 10 — x = the other. 

By the conditions, x (10 — x) = 26 



Whence, x = 5 ± y/— 1. 

Thus, we obtain an imaginary result. We therefore con- 
clude that the problem cannot be solved numerically ; in fact, 
if we call one of the parts 5 + y, the other must be 5 — y, and 
their product will be 25 — y' 2 . which, so long as y is numerical, 
is less than 25. But we are required to find two numbers 
whose sum is 10 and product 26 ; there are, then, no such 
numbers. 

Had it been required to find two expressions, whose sum is 

10 and product 25. the answer 5 + \/ — 1 and 5 — \j— 1 
would have satisfied the conditions. 

The given problem, however, expresses conditions incom- 
patible with each other, and, consequently, is impossible. 
Hence, 

Imaginary results indicate that the problem is impossible. 

PROBLEM OF THE LIGHTS. 

336. The principles of interpretation will be further illus- 
trated in the discussion of the following general problem. 

Find upon the line which joins two lights, A and B, the 
point which is equally illuminated by them : admitting that 
the intensity of a light, at a given distance, is equal to its 



2G0 ALGEBRA. 

intensity at the distance 1, divided by the square of the given 
distance. 

C" A C B C 

— I 1 J 1 1 — 

Assume A as the origin of distances, and regard all dis- 
tances estimated to the right as positive. 

Let a denote the intensity of the light A, at the distance 1 ; 
b the intensity of the light B, at the distance 1 ; and c the 
distance A B, between the two lights. 

Suppose C the point of equal illumination, and let x repre- 
sent the distance from it to A, or the distance AC. Then, 
c — x will represent the distance B C. 

By the conditions of the problem, since the intensity of the 

light A, at the distance 1, is a, at the distance x it is — = : and 

X' 

since the intensity of the light B, at the distance 1, is b, at the 

distance c — x it is -. ^ . But, by supposition, at C these 

intensities are equal ; hence, 
a b 



Whence, 



x 2 ' (c - xf ' 
c—x 



~ ( c - 


x) 2 


b 


or 

X 2 




a 


_ + v* 







x \l a 

From this equation we obtain as the values of x : 

c\J a i \J a 



x 



or, 



\Ja + ^b w \\ja + ^by 

c\J a / sj a \ 

\J a — \J b \\J a — \J b)' 



Since both a and b are positive, the two values of x are both 
real. Hence, 

There are two points of equal illumination on the line of the 
lights. 



DISCUSSION OF PROBLEMS. 261 

Since there are two lights, c must always he greater than ; 
consequently neither a, b, nor c can he 0. The problem, then, 
admits properly of only these three different suppositions : 

1. a > b. 2. a < b. 3. a=b. 

We shall now discuss the values of x under each of these 
suppositions. 

1. a > b. 

In this case, the first value of x is less than c ; because 

- being a proper fraction, is less than 1. This value 

y/ a + \J b 

c 

of x is also greater than - ; because, the denominator being 

less than twice the numerator, as b is less than a, the fraction 
is greater than \. Hence, the first point of equal illumination 
is at C, between the two lights, but nearer the lesser one. 

The second value of x is greater than c : because —A- — , 

ya — \J b 

being an improper fraction, is greater than 1. Hence, the 
second point is at C, in the prolongation of the line A B, be- 
yond the lesser light. 

These results agree with the supposition. For, if a is greater 
than b, then B evidently is the lesser light. Hence, both points 
of equal illumination will be nearer B than A ; and since the 
two lights emit rays in all directions, one of the points must 
he in the prolongation of A B beyond both lights. 

2. a < b. 

In this case, the first value of x is positive. It is also less 

than C -\ because . ^ C ' . , , having the denominator greater 
2 v a + y b 

than twice the numerator, b being greater than a, is less than 

h- Hence, the first point of equal illumination is between the 

lights, but nearer A, the lesser light, 

The second value of x is negative, because the denominator 

y/ a — y 1 b is negative ; which must he interpreted as measur- 



262 ALGEBKA. 

ing distance from A towards the left (Art. 205). Hence, the 
second point of equal illumination is at C", in the prolongation 
of the line, at the left of the lesser light, A. 

These results correspond with the supposition; the case 
being the same as the preceding one, except that A is now the 
lesser light. 

3. a = b. 
In this case, the first value of x is positive, and equal to ^ . 

Li 

Hence, the first point of equal illumination is midway be- 
tween the two lights. 

The second value of x is not finite: because -; — p- , if 

V « — V b 

I 

a = b, reduces to ~- = cc (Art. 210), which indicates that no 

finite value can be assigned to x. Hence, there is no second 
point of equal illumination in the line A B, or its prolongation. 

These results agree with the supposition. For, since the 
lights are of equal intensity, a point of equal illumination will 
obviously be midway between them ; and it is evident that 
there can be no other like point in their line. 

The preceding discussion illustrates the precision with which 
algebraic processes will conform to every allowable interpreta- 
tion of the enunciation of a problem. 



XXXI. — RATIO AND PROPORTION. 

337. The Ratio of one quantity to another of the same 
kind is the quotient arising from dividing the first quantity 
by the second (Art. 181). 

Thus, the ratio of a to b is - , or a : b. 

b 



EATIO AND PROPORTION. 263 

338. The Terms of a ratio are the two quantities required 
to form it. Of these, the first is called the antecedent, and the 
second the consequent. 

Thus, in the ratio a : b, a and b are the terms, a the ante- 
cedent, and b the consequent. 

339. A Proportion is an equality of ratios (Art. 181). 
Thus, if the ratios a : b and c : d are equal, they form a pro- 
portion, which may he written 

a : b = c : d, or a : b : : c : d. 

340. The Terms of a proportion are the four terms of its 
two ratios. The first and third terms are called the antece- 
dents ; the second and fourth, the consequents; the first and 
last, the extremes ; the second and third, the means ; and the 
terms of each ratio constitute a couplet. 

Thus, in a : b = c : d, a and c are antecedents ; b and d, con- 
sequents ; a and d, extremes; b and c, means; a and b, the 
first couplet ; and c and d, the second couplet. 

341. A Proportional is any one of the terms of a propor- 
tion ; a Mean Proportional between two quantities is either 
of the two means, when they are equal ; a Third Proportional 
to two quantities is the fourth term of a proportion, in which 
the first term is the first of the quantities, and the second and 
third terms each equal to the second quantity ; a Fourth Pro- 
portional to three quantities is the fourth term of a proportion 
whose first, second, and third terms are the three quantities 
taken in their order. 

Tims, in a : b = b : e, b is a mean proportional between a 
and c ; and c is a third proportional to a and b. In a : b — c : d, 
d is a fourth proportional to a, b, and c. 

342. A Continued Proportion is one in which each conse- 
quent is the same as the next antecedent ; as, 

a : b = b : c = c : d =■ d : e. 



264 ALGEBRA. 

PROPERTIES OF PROPORTIONS. 

343. Wlien four quantities are in proportion-, the prodwt 
of the extremes is equal to the product of the means. 

Let the proportion be 

a : b = c : d. 

This may he written (Art. 337), 

a c 
b = d 

Whence, ad = b c. 

Hence, if three quantities be in continued proportion, the 
product of the extremes is equal to the square of the means. 

Thus, if 

a : b = b : c 

then, a c = b 2 . 

By this theorem, if three terms of a proportion are given, 
the fourth may be found. Thus, if 



then, 
Whence, 

344. If the product of two quantities be equal to the prod- 
uct of two others, one pair of them may be made the extremes, 
and the other pair the means, of a proportion. 

Thus, if 

a d = b c 

ad be a c 

Dividing bv b d, -=— =■ = rr— ; , or T = -= 

& J bd bd' b d 

Whence (Art. 337), a:b = c:d. 



a : b 


= c : x 


a x 


= bc 


x 


_bc 

a 



RATIO AND PROPORTION. 265 

In a similar manner, we might derive from the equation 
a d = b c, the following proportions : 

a : c = b : c 1, 

b : d = a : c, 

c : d = a : b, 

d : b = c : a, etc. 

345. If four quantities are in proportion, they will be in 
proportion by Alternation; that is, the antecedents will 
have to each other the same ratio as the consequents. 

Thus, if a : b = c : d 

then (Art. 343), ad = bc 

Whence (Art. 344), a : c = b : d. 

346. If four quantities are in proportion, they will be in 
proportion by Inversion; that is, the second term will be to 
the first, as the fourth is to the third. 

Thus, if a : b — c : d 

then, ad — be 

Whence, b : a = d : c. 

347. If four quantities are in proportion, they will be in 
proportion by Composition; that is, the sum of the first two 
terms ivill be to the first term, as the sum of the last two terms 
is to the third term. 

Thus, if a : b — c : d 

then, ad = b c 

Adding hoth members to a c, 

ac + ad = ac + bc, or a (c ' + d) = c {a + b) 

Whence, a + b: a = c + d: c (Art. 344). 

Similarly, we may show that 

a + b : b — c + d : d. 



266 ALGEBRA. 

348. If four quantities are in proportion, they will be in 
proportion by Division; that is,, the difference of the first two 
terms will be to the first term, as the difference of the last two 
terms is to the third term. 

Thus, if a : b = c : d 

then, a d = b c 

Subtracting both members from a c, 

ac — ad = ac — be, or a (c — d)=c (a — b) 
Whence, a — b : a = c — d : c. 

Similarly, we may prove that 

a — b : b = c — d : d. 

349. If four quantities are in proportion, they will be in 
proportion by Composition and Division; that is, the sum 
of the first two terms will be to their difference, as the sum of 
the last two terms is to their difference. 



(1) 
(2) 



Thus, if 


a : b — c : d 


by Art. 347, 


a + b c + d 
a e 


and, by Art. 348, 


a — b c — d 
a c 


Dividing (1) by (2), 


a + b c + d 
a — b c — d 


or, a -f- b 


: a — b = c + d : c 



d. 

350. Quantifies which are proportional to the same quan- 
tities, are propjortional to each other. 

Thus, if a :b = e:f 

and c : d = e :f 

. ae.ee 

then, r = 3 and - 7 = ^ 

Therefore, -=- 

b d 

Whence, a:b = c:d. 



RATIO AND PROPORTION. 267 

351. If any number of quantities are proportional, any 

antecedent is to its consequent, as the sum of all the antece- 
dents is to the sum of all the consequents. 

Thus, if 

a : b = c : d = e :f 

then (Art. 343), ad = b c 

and af-=be 

also, a b = « b 

Adding, a (b + d +/) = b (a + c + e) 

Whence (Art. 344), a :b =a + c + e : b + d +f 

352. If four quantities are in proportion, if the first and 
second be multiplied or divided by any quantity, as also the 
third and fourth, the resulting quantities will be in proportion* 

Thus, if 

a : b = c : d 



then, 


a c 
b = d 


Therefore, 


ma nc 
m b n d 


Whence, 


m a : mb = n c : n d. 


In a similar 


manner we could prove 




a b c d 




m ' m n' n' 



Either m or n may be made equal to unity. That is, either 
couplet may be multiplied or divided, without multiplying or 
dividing the other. 

353. If four quantities are in proportion, and the first ami 
third be multiplied or divided by any quantity, as also the 
second and fourth, the resulting quantities will be in jiro- 
portion. 



268 ALGEBRA. 



Thus, if a : b = c : d 



then, 



Therefore, 



a c 

b = d 

m a m c 



nb n d 

Whence, m a : nb = m e : nd. 

In a similar manner we could prove 

a b _ c d 

m n m ' n 

Either m or n may he made equal to unity. 

354. If there be two sets of proportional quantities, the 
products of the corresponding terms will be in proportion. 

Thus, if a :b = c : d 

and e:f=g:h 

a c .. e q 
then, - = - and -=- 



Therefore, 



b d f h 

ae eg 



bf dh 
Whence, ae :bf= c g : d h. 

355. If four quantities are in proportion, like poioers or 
like roots of these quantities will be in proportion. 

Thus, if a :b = c :d 

then, r = --,', therefore, — = — 

b d ' b n d n 

Whence, a n :b n = c n : d n . 

In a similar manner we could prove 

y/ a : y/ b = y/ c : y d. 

356. If three quantities are in continued proportion, the 
first is to the third, as the square of the first is f>> the square 
of the second. 



RATIO AND PROPORTION. 269 



Thus, if a:b = b : c 

a b 



then, 



a 



b c 
a 2 aba 



Multiplying by -, y=l* c c 

Whence, a: c — a~ : b~. 

In a similar manner we could prove that if 

a : b = b : c = c : d, then a\d=-a z : b 3 . 

Note. The ratio a 2 : b 2 is called the duplicate ratio, and the ratio a? : b 3 
the triplicate ratio, of a : b. 

PROBLEMS. 

357. 1. The last three terms of a proportion being 18, 6, 
and 27, what is the first term ? 

2. The first, third, and fourth terms of a proportion being 
4, 20, and 55, respectively, what is the second term ? 

3. Find a fourth proportional to \, \, and \. 

4. Find a third proportional to 5 and 3. 

5. Find a mean proportional between 2 and 8. 

6. Find a mean proportional between 6 and 24. 

7. Find a mean proportional between 49 and 4. 

8. Find two numbers in tbe ratio of 2\ to 2, such that when 
each is diminished by 5, they shall be in the ratio of lj to 1. 

9. Divide 50 into two such parts that the greater increased 
by 3, shall be to the less diminished by 3, as 3 to 2. 

10. Divide 27 into two such parts that their product shall 
be to the sum of their squares as 20 to 41. 

11. There are two numbers which are to each other as 4 to 
9, and 12 is a mean proportional between them. What are 
the numbers ? 



270 ALGEBRA. 

12. The sum of two numbers is to their difference as 10 to 
3, and their product is 364. What are the numbers- ? 

13. Find two numbers such that if 3 be added to each, they 
will be in the ratio of 4 to 3 ; and if 8 be subtracted from each, 
they will be in the ratio of 9 to 4. 

14. There are two numbers whose product is 96, and the 
difference of their cubes is to the cube of their difference as 19 
to 1. What are the numbers ? 

15. Each of two vessels contains a mixture of wine and 
water ; a mixture consisting of equal measures from the two 
vessels, contains as much wine as water; and another mixture 
consisting of four measures from the first vessel and one from 
the second, is composed of wine and water in the ratio of 2 to 
3. Find the ratio of wine to water in each vessel. 

16. If the increase in the number of male and female 
criminals be 1.8 per cent, while the decrease in the number of 
males alone is 4.6 per cent, and the increase in the number of 
females alone is 9.8 per cent, compare the number of male and 
female criminals, respectively, at the first time mentioned. 

17. A railway passenger observes that a train passes him, 
moving in the opposite direction, in 2 seconds ; whereas, if it 
had been moving in the same direction with him, it would 
have passed him in 30 seconds. Compare the rates of the two 
trains. 



XXXII. — VARIATION. 

358. Variation, or general proportion, is an abridged 
method of expressing common proportion. 

Thus, if A and B be two sums of money loaned for equal 
times, at the same rate of interest, then 

A : B = A's interest : B's interest 



VARIATION. 271 

or, in an abridged form, by expressing only two terms, the in- 
terest varies as the principal ; thus (Art. 23), 

The interest oc the principal. 

359. One quantity varies directly as another, when the 
two increase or decrease together in the same ratio. 

Sometimes, for the sake of brevity, we say simply one quan- 
tity varies as another, omitting the word " directly/' 

Thus, if a man works for a certain sum per day, the amount 
of his wages varies as the number of days during which he 
works. For, as the number of days increases or decreases, the 
amount of his wages will increase or decrease, and in the same 

ratio. 

360. One quantity varies inversely as another, when the 
first varies as the reciprocal of the second. 

Thus, if a courier has a fixed route, the time in which he 
will pass over it varies inversely as his speed. That is, if he 
double his speed, he will go in half the time; and so on. 

361. One quantity varies as two others jointly, when it 
has a constant ratio to the product of the other two. 

Thus, the wages of a workman will vary as the number -of 
days he has worked, and the wages per day, jointly. 

362. One quantity varies directly as a second and inversely 
as a third, when it varies jointly as the second and the recip- 
rocal of the third. 

Thus, in physics, the attraction of a planetary body varies 
directly as the quantity of matter, and inversely as the square 
of the distance. 

363. If A varies as B, then A is equal to B multiplied by 
some constant quantity. 

Let a and b denote one pair of corresponding values of two 
quantities, and A and B any other pair. Then, by Art. 358, 



272 ALGEBRA. 

A:a = B:b 

Whence (Art. 343), Ab = a B, or A=^B 

• a 
Denoting the constant ratio - by m, 

A = m B. 

Hence, also, when any quantity varies as another, if any 
two pairs of values of the quantities be taken, the four will be 
proportional. 

For, if A oc B, and. A' and B' be any pair of values of A and 
B, and A" and B" any other pair, by Art. 363, 

A' = m B', and A" = m B" 



a 



A 1 A 

Whence, -~ = m, and — - 



Therefore, 



A" 



B 1 B" 

or (Art. 337), A':B' = A" : B 



'ii 



364. The terms used in Variation may now be distin- 
guished as follows : 

1. If A = m B, A varies directly as B. 

Ml 

2. If A = — , A varies inversely as B. 

3. If A = m B C, A varies jointly as B and C. 

4. If A — , A varies directly as B, and inversely as C. 

Problems in variation, in general, arc readily solved by con- 
verting the variation into an equation, by the aid of Art. 364. 



VAKIATION. 273 

EXAMPLES. 

365. 1. Given that y oc x, and when x = 2, y = 10. Re- 
quired the value of y in terms of x. 

If y oc x, by Art. 364, y = mx 

Substituting x = 2 and y = 10, 10 = 2 m, whence m = 5. 
Hence, the required value is y = 5 a?. 

2. Given that 7/ oc »■, and that v/ = 2 when x = l. What 
will he the value of y when x = 2 ? 

3. If y oc £, and y = 21 when £ = 3, find the value of y in 
terms of z. 

4. If x varies inversely as y, and x = 4 when y = 2, what is 
the value of cc when y = 6? 

5. Given that z varies jointly as x and y, and that z = 1 
when x = 1 and ?/ = 1. Find the value of z when x = 2 and 
y = 2. 

6. If ?/ equals the sum of two quantities, of which one is 
constant, and the other varies as x y ; and when x = 2, y = — 2 J-, 
hut when x = — 2, y = 1 ; express y in terms of x. 

7. Two circular plates of gold, each an inch thick, the diam- 
eters of which are 6 inches and 8 inches, respectively, are 
melted and formed into a single circular plate one inch thick- 
rind its diameter, having given that the area of a circle varies 
as the square of its diameter. 

8. Given that the illumination from a source of light varies 
inversely as the square of the distance ; how much farther from 
a candle must a book, which is now 3 inches off, be removed, 
so as to receive just half as much light ? 

' 9. A locomotive engine without a train can go 24 miles an 
hour, and its speed is diminished by a quantity which varies 
as the square root of the number of cars attached. With four 
cars its speed is 20 miles an hour. Find the greatest number 
of cars which the engine can move. 



274 ALGEBEA. 

XXXIII. — ARITHMETICAL PROGRESSION. 

366. An Arithmetical Progression is a series of quanti- 
ties, in which each term is derived from the preceding term 
by adding a constant quantity, called the com raon difference. 

367. When the series is increasing, as, for example, 

1,3,5,7,9,11, 

each term is derived from the preceding term by adding a 
positive quantity; consequently the common difference is 
positive. 

When the series is decreasing, as, for example, 

19, 17,15,13,11,9, 

each term is derived from the preceding term by adding a 
negative quantity; consequently the common difference is 
negative. 

368. Given the first term, a, the common difference, d, and 
the number of terms, n, to find the last term, I. 

The progression will be 

a, a + d, a + 2 d, a + 3 d, 

We observe that these terms differ only in the coefficient of 
d, which is 1 in the second term, 2 in the third term, 3 in the 
fourth term, etc. Consequently in the rath term, the coefficient 
of d will be n — 1. Hence, the rath term of the series, or the 
last term, as the number of the terms is n, will be 

l = a+ (n-l)d (1) 

369. Given the first term, a, the last term, 1, and the num- 
ber of terms, n, to find the sum of the series, S. 

S=a+ (a + d) + (a + 2d) + + (l-2d) + (l — d) + l 

Writing the terms of the second member in the reverse 
order, 

S=l+ (l-d) + (I- 2 d) + + {a + 2 d) + (a + d) + a 



ARITHMETICAL PROGRESSION. 275 

Adding these equations, term by term, we have 
2S=(a+l) + (a + l) + (a + l)+ + (a+t)+(a+t) + (a+l) 

In this result, (a + I) is taken as many times as there are 
terms, or n times ; hence 

2S=n(a+l),ovS= 7i (a+I) (2) 

Using the value of I given in (1), Art, 3G8, this may he 
written 

S=%[2a + (n-l)d] 

370. 1. In the series 5, 8/11, to 18 terms, find the 

last term and the sum of the series. 

Here a = 5, n = 18 ; the common difference is always found 
by subtracting the first term from the second; hence 
d = 8-5 = 3. 

Substituting these values in (1) and (2), we have 

I = 5 + (18 - 1) 3 = 5 + 17 x 3 = 5 + 51 = 56. 

1 S 
S = -^ (5 + 56) = 9 x 61 = 549. 

2. In the series 2,-1,-4, to 27 terms, find the last 

term and the sum of the series. 

Here a = 2, n = 27, d = the second term minus the first 
= — 1 — 2 = — 3. Substituting these values in (1) and (2), 
we have 

1 = 2 + (27-1) (-3) = 2 + 26 (-3) =2- 78 = -76. 

S=^-(2-76)=^-(-74) = 27(-37) = -999. 

« 

EXAMPLES. 

Find the last term and the sum of the series in the fol- 
lowing : 



276 ALGEBRA. 

3. 1, 6, 11, to 15 terms. 

4. 7, 3, — 1, to 20 terms. 

5. — 9, — 6,-3, to 23 terms. 

6. -5,-10,-15, to 29 terms. 

^ / o o . 

o> 7> k> to 16 terms. 

8. ■=, T-p , to 19 terms. 

5 15 

1 5 

9. - , — , to 22 terms. 

2 1 

10. — - , - , to 14 terms. 

o o 

5 

11. — 3, — -, to 17 terms. 

113 

12. j j 9 j j 3 to 35 terms. 

371. Formulae (1) and (2) constitute two independent 
equations, together containing all the five elements of an 
arithmetical progression ; hence, when any three of the five 
elements are given, we may readily deduce the values of the 
other two, as by substituting the three known values we shall 
have two equations with only two unknown quantities, which 
may be solved by methods previously given. 

1. The first term of an arithmetical progression is 3, the 
number of terms 20, and the sum of the terms 440. Find the 
last term and the common difference. 

Here a = 3, n = 20, £=440; substituting in (1) and (2), 
we have 

1 = 3 + 19 d 

- 440 = 10 (3 + I), or 44 = 3 + I 

From the second equation, I = 41 ; substitute in the first, 

41 = 3 + 19 d ; 19 d = 38 ; d = 2. 



ARITHMETICAL PROGRESSION. 277 

2. Given d = — 3, I = — 39, S = — 264 ; find a and n. 
Substituting the given quantities in (1) and (2), 

- 39 = a + (n - 1) (- 3), or 3 n - a = 42 

- 264 = - (a - 39), or a n — 39 n = - 528 

From the first of these equations, a — 3 ?i — 42 ; substitute 
in the second, 

(3 n - 42) n - 39 » = - 528, or ?r - 27 n = - 176 

Whence, n = - ~ = ^—^ — = 16 or 11 

Substituting in the equation a = 3n — 42, 

When ii — 16, a = 6 

n = 11, a. = — 9, Ans. 

The signification of the two answers is as follows : 
If n = 16, and a = 6, the series will be 

6, 3, 0, - 3, - 6, - 9, - 12, - 15, - 18, - 21, - 24, - 27, 

- 30, - 33, - 36, - 39. 

If n = 11, and a = — 9, the series will be 
_ o, - 12, - 15, - 18, - 21, - 24, - 27, - 30, - 33, - 36, - 39. 
In either of which the last term is — 39 and the sum — 264. 

113 

3. Given a = ^ , d — — T x, S = — ^; find I and n. 

Substituting the given quantities in (1) and (2), we have 
J = |+(»_l)(-l),or 12l + n*=5 



S_nfl 

'2~2 



(= + l) , ox n + 3 1 n = — 9 



278 ALGEBRA. 

From, the first of these, n = 5 — 12 I ; substitute in the 
second, 

5 - 12 1 + 3 I (5 - 12 I) = - 9, or 36 V" - 3 1 = 14 

, 3 ±y/ 9 + 2016 3 + 45 2 7 

Whence, J = ^ = 70 = 3 ° r ~~ 12 



Substituting in the equation re = 5 — 12 £, 



9 



"When I = o , n — — 3 



o 



1 = — — , n = 12, Ans. 

The first answer is inapplicable, as a negative number of 
terms has no meaning. Hence the only answer is, 

7 
l = — j2> ft = 12. 



Note. A negative or fractional value of n is always inapplicable, and 
should be neglected, together with all other values dependent on it. 



EXAMPLES. 

4. Given d = 4, 1 = 75, re = 19 ; find a and >S f . 

165 

5. Given cZ = — 1, re = 15, $ = — - ; find a and Z. 

-j 

2 

6. Given a = — -, re = 18, £ = 5 ; find d and & 

o 

3 

7. Given a = — — , re = 7, $ = — 7 ; find d and Z. 

8. Given I = — 31, re = 13, S= - 169 ; find a and tf. 

9. Given a = 3, I = 42-f , d = 2£ ; find re and S. 

10. Given a = 7, l = — 73, #=— 363 ; find re and c?. 

n n- 15 5 2625 

11. Given a = — , d = 7 :, o= ; find re and Z. 

£ Ji Ji 



ARITHMETICAL PROGRESSION. 279 

12. Given I = - 47, d = - 1, 8= - 1118 ; find a and n. 

13. Given d = — 3, S — — 328, a = 2; find Z and n. 

372. 2b insert any number of arithmetical means between, 
two given terms. 

1. Insert 5 arithmetical means between 3 and — 5. 

This may he performed in the same manner as the examples 
in the previous article ; we have given the first term a = 3 ; 
the last term I = — 5 ; the number of terms n = 7 ; as there 
are 5 means and two extremes, or in all 7 terms. Substituting 
in (1), Art. 368, we have 

4 

— 5 = 3 + 6 d ; or, 6 d = — 8 ; whence, d = — -= . 

o 

4 
Hence the terms are obtained by subtracting - from 3 for 

4 

the first, - from that result for the second, and so on ; or, 

3 5 1 1 -l - 11 -5 Ans 
«5j o i o j — L t — o j o" > °> sins. 

EXAMPLES. 

2. Insert 5 arithmetical means between 2 and 4. 

3. Insert 7 arithmetical means between 3 and — 1. 

4. Insert 4 arithmetical means between — 1 and — 6. 

5. Insert 6 arithmetical means between — 8 and — 4. 

6. Insert 4 arithmetical means between — 2 and 6. 

7. Insert m arithmetical means between a and b. 

PROBLEMS. 

373. 1. The 6th term of an arithmetical progression is 19, 
and the 14th term is 67. Find the first term. 

By Art. 368, the 6th term is a + 5 d, and the 14th term is 
a + 13 d. Hence, 



280 ALGEBRA. 

a+ 5d = 19 
a ±13 d = 67 



Whence, 8 d = 48, or d = 6 

Therefore, a = — 11, Ans. 

2. Find four quantities in arithmetical progression, such 
that the product of the extremes shall he 45, and the product 
of the means 77. 

Let a, a + d, a + 2 d, and a ± 3 d he the quantities. 
Then, by the conditions, a 2 ± 3 a d = 45 (1) 

a 2 ± 3 a d + 2 d 2 = 77 (2) 

Subtracting (1) from (2), 2 d 2 = 32 

d 2 = 16 
t?=±4. 
If d = 4, substituting in (1), we have 

a 2 + 12a = 45 

_ -12±V^Ti4TT80 -12 ±18 Q 
Whence, a = —^ = ^ = 3 or — 15. 

This indicates two answers, 

3, 7, 11, and 15, or, — 15, — 11, — 7, and — 3. 
If d = — 4, substituting in (1), we have 

a 2 — 12 a- 45 



W1 12±V144 + 180 12 ±18 1K Q 

Whence, a = ! — ^ = „ = 15 or — 3. 

This also indicates two answers, 

15, 11, 7, and 3, or, — 3, — 7, — 11, and — 15. 

But these two answers are the same as those obtained with 
the other value of d. Hence, the two answers to the problem 
are 

3, 7, 11, and 15, or, — 3, — 7, — 11, and — 15. 



ARITHMETICAL PROGRESSION. 281 

3. Find the sum of the odd numbers from 1 to 100. 

4. A debt can be discharged in a year by paying $ 1 the 
first week, $ 3 the second, $ 5 the third, and so on. .Required 
the last payment, and the amount of the debt. 

5. A person saves $270 the first year, $210 the second, 
and so on. In how many years will a person who saves every 
year $ 180 have saved as much as he ? 

6. Two persons start together. One travels ten leagues a 
day, the other eight leagues the first day, which he augments 
daily by half a league. After how many days, and at what 
distance from the point of departure, will they come together ? 

7. Find four numbers in arithmetical progression, such that 
the sum of the first and third shall be 22, and the sum of the 
second and fourth 36. 

8. The 7th term of an arithmetical progression is 27 ; and 
the 13th term is 51. Find the first term. 

9. A gentleman set out from Boston to NeW York. He 
travelled 25 miles the first day, 20 miles the second day, each 
day travelling 5 miles less than on the preceding. How far 
was he from Boston at the end of the eleventh day ? 

10. If a man travel 20 miles the first day, 15 miles the sec- 
ond, and so continue to travel 5 miles less each day, how far 
will he have advanced on his journey at the end of the 8th 
day? 

11. The sum of the squares of the extremes of four quanti- 
ties in arithmetical progression is 200, and the sum of the 
squares of the means is 13G. What are the quantities ? 

12. After A had travelled for 2| hours, at the rate of 4 
miles an hour, B set out to overtake him, and went 4£ miles 
the first hour, 4J the second, 5 the third, and so on, increasing 
his speed a quarter of a mile every hour. In how many' hours 
would he overtake A ? 



282 ALGEBRA. 



XXXIV. — GEOMETRICAL PROGRESSION. 

374. A Geometrical Progression is a series in which each 
term is derived from the preceding term by multiplying by a 
constant quantity, called the ratio. 

375. When the series is increasing, as, for exanrple. 

2, 6, 18, 54, 162, 

each term is derived from the preceding term by multiplying 
by a quantity greater than 1 ; consequently the ratio is a 
quantity greater than 1. 

When the series is decreasing, as, for example, 
9 3 1 i i jl 

each term is derived from the preceding term by multiplying 
by a quantity less than 1 ; consequently the ratio is a quantity 
less than 1. 

Negative values of the ratio are admissible ; for example, 

-3, 6,-12,24,-48, 

is a progression in which the ratio is — 2. 

376. Given the first term, a, the ratio, r, and the number 
of terms, n, to find the last term, I. 

The progression will be 

a, ar, ar 2 , ar 5 , 

We observe that the terms differ only in the exponent of r, 
which is 1 in the second term, 2 in the third term, 3 in the 
fourth term, etc. Consequently in the nth. or last term, the 
exponent of r will be n — 1, or 

l=ar n - x (1) 

377. Given the first term, a, the last term, I, and the ratio. 
r, to find the sum of the series, S. 



GEOMETRICAL PROGRESSION. 283 

S= a + a r + a r 2 + a r 3 + + a r n ~ 3 + a r"- 2 + a r n ~ x 

Multiplying each term by r, 

r S—ar+ ar 2 +ar 3 + a7 A + + a r n ~ 2 + a r" _1 + a r n 

Subtracting the first equation from the second, we have 

ft 2 ,Tl (f 

r S— S=ar n — a, or S (r —l) — a r n — a, or S = - — ^— 
But from (1), Art. 376, by multiplying each term by r, 

Substituting this value of a r n in the value of S, we have 

r — 1 

378. 1. In the series 2, 4, 8, to 11 terms, find the last 

term and the sum of the series. 

Here a = 2, n = ll; the ratio is always found by dividing 

4 

the second term by the first ; hence, r = - = 2. 

Substituting these values in (1) and (2), we have 
I = 2 (2) 11 - 1 = 2 x 2 10 = 2 x 1024 = 2048. 
s= (2x204S)-2 =4096 _ 2 = 4094 

Zi — JL 

2. In the series 3, 1. », to 7 terms, find the last term 

o 

and the sum of the series. 

Here a = 3, n = 7, r = second term divided by first term = ^. 

Substituting these values in (1) and (2), we have 
l - 6 \3) ~* \3) _ 3 6_ 3 5_ 243" 



284 ALGEBEA. 

a j_v , r , 2186 

v3 X 243/ ° _729 " 729 _ 2186 3 _ 1093 

^ = ~^[ ~ - 1 — " ^2~ : T29" X 2 ~" ^43" ' 

3 _1 3 ~3 

3. In the series — 2, 6, — 18, to 8 terms, find the last 

term and the sum of the series. 

Here a — — 2, n = S, r= — ~ = — 3. Hence, 

I = (_ 2) (- 3) 8 - 1 = (- 2) (- 3) 7 = (- 2) (- 2187) = 4374. 

(_ 3 x 4374) - (- 2) - 13122 + 2 _^ - 13120 = 
(-3) — 1- -4 -4 

EXAMPLES. 

Find the last term and the sum of the series in the fol- 
lowing : 

4. 1, 2, 4, to 12 terms. 

4 

5. 3, 2, - , to 7 terms. 

o 

6. —2, 8, —32, to 6 terms. 

7. 2, — 1, ■= , to 10 terms. 



111 

2' V 8 



8. ^ , T , q , to 11 terms. 



2 3 
9. k i — 1? ^ i to 8 terms. 

10. 8, 4, 2, to 9 terms. 

11. 4' -4' J2' to6 termS * 

2 11 

12. -k>— g>— gj to 10 terms. 

13. 3, -6, 12, to 7 terms. 



GEOMETRICAL PROGRESSION. 285 

379. Formulae (1) and (2) together contain all the five ele- 
ments of a geometrical progression ; hence, if any three of the 
five are given, we may find the other two, exactly as in arith- 
metical progression. But in certain cases the operation in- 
volves the solution of an equation of a higher degree than the 
second, for which rules have not heen given ; and in other 
cases the unknown quantity appears as an exponent, the solu- 
tion of which equations can usually only be effected by the use 
of logarithms ; although in certain simple cases they may be 
solved by inspection. 

1. Given I = 6561, r — 3, n = 9 ; find a and S. 

Substituting these values in (1) and (2), Arts. 376 and 377, 
we have 

6561 = a (3) 8 ; or 6561 = 6561 a ; or a = 1. 

(3x6561)-l_ 19683-1 _ 19682 _ 
6 ~ 3=1 ~" ~2~ "" ~2~ 

2. Given a = — 2, n = 5, I = — 32 ; find r and S. 
Substituting these values in (1) and (2), we have 

-32 = (-2)(r) 5 - 1 ; or-32 = -2r 1 ; r 4 = 16; r = ±2. 

If r = 2, S= (2 X ~ 32) 7 ( ~ 2) =- 64 + 2 = -62. 

T , 9 q (-2x-32)-(-2) _ 64 + 2 _66_ 

The signification of the two answers is as follows : 

If r=2, the series will be -2, -4, -8, -16, -32, in 
which the sum is — 62. 

If r = — 2, the series will be — 2, 4, — 8, 16, — 32, in which 
the sum is — 22. 

3. Given a = 3, r = — ^ , S = ; find n and I. 



286 ALGEBRA. 

Substituting these values in (1) and (2), we have 

-Us 

1640 _ 3 1+9 • 6560 1 

"729— ^TV=^r ; oW+9 = T29 ; ° Tl = -729- 
3 

3 

Substituting this value of I in the equation (— 3)" _1 = -, we 

3 
have (— 3) n ~ 1 = ^- = — 2187; whence, by inspection, n— 1 



71".) 
= 7, or n = 8. 

EXAMPLES. 

4. Given J = - 256, r = — 2, n = 10 ; find a and A 

5. Given r = -, n = 8, S— -^-^r ; find a and Z. 

o 6561 

6. Given a = 2, n = 7, I = 1458 ; find r and 5. 

3 

7. Given a = 3, « = 6, Z = — . ,^ , ; find r and & 

1024 

8. Given a-= 1, r = 3, Z = 81 ; find -« and S. 

1 127 

9. Given a = 2, Z = ^ , $ = -^- ; find n and r. 

10. Given «. = - , r = — 3, $= — 91; find n and Z. 

11. Given £ = -128, r = 2, £ = -255; find n and a. 

380. The Limit to which the sum of the terms of a decreas- 
ing geometrical progression approaches, as the number of terms 
becomes larger and larger, is called the sum of the scries to 
infinity. We may write the value of S obtained in Art. .'177 
as follows : 



GEOMETRICAL PROGRESSION. 287 

a — r I 



S-. 



1-r 



In a decreasing geometrical progression, the larger the num- 
ber of terms taken the smaller will be the value of the last 
term ; hence, by taking terms enough, the last term may be 
made as small as we please. Then (Art. 207), the limiting 
value of I is 0. Consequently the limit to which the value of 
S approaches, as the number of terms becomes larger and 

larger, is - . 

Therefore the sum of a decreasing geometrical progression 
to infinity is given by the formula 

1. Find the sum of the series 3, 1, ^ , to infinity. 

o 

Here a = 3, r = k ; substituting in (3), we have 

O 3 9 9 A 

3 

8 16 

2. Find the sum of the series 4, — ^ , — , to infinity. 

_8 

~~ 3 2 
Here a = 4, r = —r- = — » ; substituting in (3), we have 

+ 3 

EXAMPLES. 

Find the sum of the following to infinity : 

3l 2 ' lj 2' 5 - - 1 '3'~9' 



288 


ALGEBRA. 


3 11 

'■ 4' 2' 3' '■" 


9 8 2 1 
9 - *'5' 50' 



8 2 -A li in -4 4 _A 

5' 35' 245' ' 5' " 25' 

381. To find the value of a rejoeating decimal. 

This is a case of finding the sum of a geometrical progres- 
sion to infinity, and may be solved by the formula of Art. 380. 

1. Find the value of .363636 

.363636 = .36 + .0036 + .000036 + 



AAO/> 

Here a = .36, and r = '—— — = .01 ; substituting in (3), 

.oo 

.36 _ : 36_36_ £ 

2. Find the value of .285151 

.285151 = .28 + .0051 + .000051 + 



To find the sum of all the terms except the first, we have 
a = .0051, r = .01 ; substituting in (3), 

a .0051 .0051 51 17 

o = 



1-.01 .99 9900 3300 

Adding the first term to this, the value of the given decimal 

28 17 941 
~ 100 + 3300" 3300' 



EXAMPLES. 

Find the values of the following : 

3. .074074 5. .7333 7. .113333. 

4. .481481 6. .52121 8. .215454. 



GEOMETRICAL PROGRESSION. 289 

382. To insert any number of geometrical means between 
two given terms. 

64 

1. Insert 4 geometrical means between 2 and p-r^ . 

This may be performed in the same manner as the examples 

64 
in Art. 379. We have a = 2, I = —-^ , and n = 6, or two more 

243 

than the number of means. 

Substituting these values in (1), Art. 376, we have 

64 32 2 

243 = 2? ' 5; ° rr5 = 243 ; mr = t 

2 
Hence the terms are obtained by multiplying 2 by ^ for the 

2 
first, that result by ^ for the second, and so on ; or, 
o 

4 8 16 32 ^4 
2 '3' 9' 27' 81' 243' 

2. Insert 5 geometrical means between — 2 and — 128. 

Here a = — 2, I = — 128, n = 7. Substituting in (1), Art. 
376, we have 

— 128 = — 2 7* ; or r 6 = 64 ; whence, r = ± 2. 

If r = 2, the series will be 

_ 2; -4, -8, -16, -32, -64, -128. 

If r = — 2, the series will be 

- 2, 4, - 8, 16, - 32, 64, - 128. 



EXAMPLES. 

o 1 128 

3. Insert 6 geometrical means between o and ^Hq • 

4. Insert 5 geometrical means between ^ and 364£. 

5. Insert 6 geometrical means between — 2 and — 4374. 



200 ALGEBRA. 

729 
6. Insert 4 geometrical means between 3 and — 



3 



1024 



7. Insert 7 geometrical means between - and 



PROBLEMS. 

383. 1. What is the first term of a geometrical progression, 
when the 5th term is 48, and the 8th term is — 384 ? 

By Art. 376, the 5th term is a r 4 , and the 8th term is a r~. 

Hence, 

ar* = 48, and a r' = — 384. 

Dividing the second of these equations by the first, 
r 3 = — 8 ; whence, r = — 2. 

Tl 48 48 <* J 

Ihen, « = — ; - = :r7r= z: 3 , Ans. 

r* 16 

2. Find three numbers in geometrical progression, such that 
their sum shall be 14, and the sum of their squares 84. 

Let "a, a r, and a r~ denote the numbers. Then, by the con- 
ditions, 

a + a r + a r 2 = 14 (1) 

a 2 + a 2 r 2 + a 2 r i = 8A (2) 

Dividing (2) by (1), a — ar+ar 2 = 6 (3) 

Adding (1) and (3), a + a r 2 = 10 (4) 

4 

Subtracting (3) from (1), a r = 4, ofr = - (5) 

1 c 

Substituting from (5) in (4), a -\ = 10 



<r 



-10 a = -16 



wi (K onox lOiy/100-64 10 ±6 Q _ 
VV hence (Art. 309), a — ~ - = — - — — 8 or 2. 

— Z 

4 1 
If a = 8, r = q = jr, and the numbers are 8, 4, and 2. 

8 Z 



HARMONICAL PROGRESSION. 291 

4 
If a = 2, ?• = - = 2, and the numbers are 2, 4, and 8. 

Therefore, the numbers are 2, 4, and 8, Ans. 

3. A person who saved every year half as much again as he 
saved the previous year, had in seven years saved $2059. 
How much did he save the first year ? 

4. A gentleman boarded 9 days, paying 3 cents for the first 
day, 9 cents for the second day, 27 cents for the third day, and 
so on. Required the cost. 

5. Suppose the elastic power of a ball that falls from a 
height of 100 feet, to be such as to cause it to rise 0.9375 of 
the height from which it fell, and to continue in this way 
diminishing the height to which it will rise, in geometrical 
progression, till it comes to rest. How far will it have moved ? 

6. The sum of the first and second of four quantities in 
geometrical progression is 15, and the sum of the third and 
fourth is 60. Required the quantities. 

7. The fifth term of a geometrical progression is — 324, and 
the 9th term is — 26244. What is the first term ? 

8. The third term of a geometrical progression is ^-r , and 

9 
the sixth term is -^r^ . What is the second term ? 



XXXV.— HARMONICAL PROGRESSION. 

384. Quantities are said to be in Harmonical Progression 
when their reciprocals form an arithmetical progression. 

For example, 1, -, F , -, 



3' 5' 7 
are in harmonical progression, because their reciprocals, 

1, 3, 5, 7, 

form an arithmetical progression. 



292 ALGEBRA. 

385. From the preceding it follows that all problems in 
harmonical progression, which are susceptible of solution, may 
be solved by inverting the terms and applying the rules of the 
arithmetical progression. There will be found, however, no 
general expression for the sum of a harmonical series. 

386. To find the last term of a given ha rmonical series. 



2 2 

3' 5 ; 



1. In the series 2, - , p, to 36 terms, find the last term. 



Inverting the series, we have the arithmetical progression 

2 j 2 ' 2 ' t0 36 terms • 

Here a = -, d = l, w = 36 ; hence, by (1), Art. 368, 

Z = i+(36-l)l = ^ + 35 = ^. 

2 
Inverting this, we obtain -=r as the last term of the given series. 



EXAMPLES. 

Find the last terms of the following : 

K Q A *} 1 9 

2. - , - , to 23 terms. 4. r,T,-, to 26 terms. 

3. - ---r, to 17 terms. 5. a, b, to n terms. 

2 3 o 

387. To ii/si'rf any number of harmonical means between 
two given terms. 

1. Insert 5 harmonical means between 2 and — 3. 
Inverting, we have to insert 5 arithmetical means between 

1 A 1 

2 and ~3' 



HARMONICAL PROGRESSION. 293 

Here a = - , Z = — -,n = J; substituting in (1), Art, 368, 
Z o 

we have 

— - — - + 6d: or 6 d — — -x : whence, cZ = — — . 
3 2 6 ob 

Hence, the arithmetical means are 

13 2 1_ _1_ _L 
36' 9' 12' 18' 36* 

Then, the harmonical means will be 

36 9 ._ 1Q 36 

-,-,12,-lS, - T ,Ans. 

EXAMPLES. 

2 3 

2. Insert 7 harmonical means between - and — . 

5 10 

3. Insert 3 harmonical means between — 1 and — 5. 

4. Insert 6 harmonical means between 3 and — 1. 

5. Insert m harmonical means between a and b. 

388. If three consecutive terms of a harmonical progres- 
sion be taken, the first has the same ratio to the third, that 
the first minus the second has to the second minus the third. 

Let a, b, c be in harmonical progression ; then their recip- 
rocals - , - , and - will be in arithmetical progression. Hence^ 
a> o c 

1_1_1_1 

c b b a' 

Clearing of fractions, ab — ac = ac — bc 
or, a (b — c) = c (a — b) 

Dividing through by c (b — c), we have 

a a — b 

which was to be proved. 



294 » ALGEBRA. 

389. Let a and c be any two quantities ; b their harrnoni- 
cal mean. Then, by the previous theorem, - = . 

Clearing of fractions, ab — ac = ac — be; then, ab + bc = 2ac 

2 a c 
or, b = . 

a + c 

390. We may note the following results : if a and c arc 

a ~f~ c 
any two quantities, their arithmetical mean = — - — ; their 

geometrical mean = \a c ; and their harmonical mean = - - . 

a+ c 

a- 2 a c a + c / / — \2 . 

Since X — 7. — = vV a c ) > l t follows that the product 

a -\- c ~j 

of the harmonical and arithmetical means of two quantities is 

equal to the square of their geometrical mean. 

Consequently the geometrical mean must be intermediate in 

value between the harmonical and the arithmetical mean. But 

the harmonical mean is less than the arithmetical mean, be- 

a + c 2ac (a + e) 2 — 4 a c a 2 + 2 ac+ c 2 — £ac 

cause — pr = - — T —/- r = — — r 

2 a + c 2 (a + c) 2 (a + c) 

a 2 — 2ac + c 2 (a — c) 2 

= — 7T7 ^ = 7T^ n a positive quantity. 

2 (a + c) 2 (a + c) ' L >■ * 

Hence of the three quantities, the arithmetical mean is the 
greatest, the geometrical mean next, and the harmonical mean 
the least. 



XXXVI. — PERMUTATIONS AND COMBINA- 
TIONS. 

391. The different orders in which quantities can be ar- 
ranged are called their Permutations. 

Thus, the permutations of the quantities a, b, <; taken two 
at a time, are 



PERMUTATIONS AND COMBINATIONS. 295 

a b, b a ; a c, c a; be, c b; 
and taken three at a time, are 

a b c, ac b; b a e, b c a ; cab, cb a. 

392. The Combinations of quantities are the different col- 
lections that can he formed out of them, without regard to the 
order in which they are placed. 

Thus, the combinations of the quantities a, b, c, taken two 

at a time, are 

a b, ac, be; 

a b, and b a, though different permutations, forming the same 
combination. 

393. To find the number of permutations of n quantities, 

taken r at a time. 

Let P denote the number of permutations of n quantities, 
taken r at a time. By placing before each of these the other 
n — r quantities one at' a time, we shall evidently form P (n — r) 
permutations of the n quantities, taken r + 1 at a time. That 
is, the number of permutations of n quantities, taken r at a 
time, multiplied by n — r, gives the number of j)ermutations 
of the n quantities, taken r + 1 at a time. 

But the number of permutations of n quantities, taken one 
at a time, is obviously n. Hence, the number of permutations 
taken two at a time, is the number taken one at a time, 
multiplied by n — 1, or n (n— 1). The number of permuta- 
tions, taken three at a time, is the number taken two at a time, 
multiplied by n — 2, or n (n — 1) (n — 2); and so on. We 
observe that the last factor in the number of permutations is 
n, minus a number 1 less than the number of quantities taken 
at a tinie. Hence, the number of permutations of n quanti- 
ties, taken r at a time, is 

n(n-l) 0-2) (n-(r-l)) 

or, n(n — l)(n — 2) (n — r + T). (1) 



296 ALGEBRA. 

394. If all the quantities are taken together, r = n and 
Formula (1) becomes 

n(n-l) (n-2) 1; 

or, by inverting the order of the factors, 

1x2x3 (n— l)n. (2) 

That is, 

The number of permutations of n quantities, taken n at a 
time, is equal to the product of the natural numbers from 1 
up to n. 

For the sake of brevity, this result is often denoted by \n, 
read "factorial n " ; thus \n_ denotes the product of the natu- 
ral numbers from 1 to n inclusive. 

395. To find the number of combinations of n quantities, 
taken r at a time. 

The number of permutations of n quantities, taken r at a 
time, is (Art. 393), 

n{n — l) (n — 2) (n — r+ 1). 

By Art. 394, each combination of r quantities produces \r 

permutations. Hence, the number of combinations must equal 
the number of permutations divided by ta or 

n(n-l) (n-2) (n-r+1) 

| r 

396. The number of combinations of n quant It Its, taken r 
at a time is the same as the number of combinations of n 
quantities taken n — r at a time. 

For, it is evident that for every combination of r quantities 
which we take out of n quantities, we leave one combination 
of n — r quantities, which contains the remaining quantities. 

EXAMPLES. 

397. 1. How many changes can be rung with 10 bells, 
taking 7 at a time ? 



PERMUTATIONS AND COMBINATIONS. 2! >7 

Here, n = 10, r = 7 ; then n — r + 1 = 4. 

Then, by Formula (1), 

10x9x8x7x6x5x4 = 604800, ^».s. 

2. How many different combinations can be made with 5 
letters out of 8 ? 

Here, n = 8, r = 5 ; then n — r + 1 = 4:. 

Then, by Formula (3), 

8x7x6x5x4 



1x2x3x4x5 



56, Ans. 



3. In how many different orders may 7 persons be seated 
at table ? 

Here n = 7 ; then, by Formula (2), 

1x2x3x4x5x6x7 = 5040, Ans. 

4. How many different words of 4 letters each can be made 
with 6 letters ? How many words of 3 letters each ? How 
many of 6 letters each ? How many in all possible ways ? 

5. How often can 4 students change their places in a class 
of 8, so as not to preserve the same order ? 

6. From a company of 40 soldiers, how many different pick- 
ets of 6 men can be taken ? 

7. How many permutations can be formed of the 26 letters 
of the alphabet, taken 4 at a time ? 

8. How many different numbers can be formed with the 
digits 1, 2, 3, 4, 5, 6, 7, 8, 9, taking 5 at a time, each digit 
occurring not more than once in any number ? 

9. How many different permutations may be formed of the 
letters in the word since, taken all together ? 

10. How many different combinations may be formed of the 
letters in the word forming, taken three at a time ? 



298 ALGEBRA. 

11. How many different combinations may be formed of 20 
letters, taken 5 at a time ? 

12. How many different combinations may be formed of 18 
letters, taken 11 at a time ? 

13. How many different committees, consisting of 7 persons 
each, can be formed out of a corporation of 20 persons ? 

14. How many different numbers, of three different figures 
each, can be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, ? 



XXXVII. — BINOMIAL THEOREM. 

POSITIVE INTEGRAL EXPONENT. 

398. The Binomial Theorem, discovered by Newton, is a 
formula, by means of which any binomial may be raised to any 
required power, without going through the process of invo- 
lution. 

399. To prove the Theorem for a positive integral ex- 
ponent. 

By actual multiplication we may show that 
(a + x) 2 = a 2 + 2 ax + x 2 
(a + x) 3 = a 3 + 3 a 2 x + 3 a x 2 + x 3 
(a + x) 4 = a 4 + 4 a 3 x + 6 a 2 x 2 + 4 a x 3 + x* 

In these results we observe the following laws : 

1. The number of terms is one more than the exponent of the 
binomial. 

2. The exponent of a in the first term is the same as the 
exponent of the binomial, and decreases by one in each stic- 
ceedin;/ term. 

3. The exponent of x in the second term is unity, and in- 
creases by one in each succeeding term. 



BINOMIAL THEOREM. 299 

4. The coefficient of the first term, is unity ; and of the sec- 
ond term, is the exponent of the binomial. 

5. If the coefficient of any term be multiplied by the expo- 
nent of a in that term, and the product divided by the number 
of the term, beginning at the left, the result will be the co- 
efficient of the next term,. 

Assuming that the laws hold for any positive integral expo- 
nent, n r we have 

. . , n(n— 1) „ „ n(n— T)(n— 2) „ , 
(a+x) n =a n +na n - 1 x+ \ ' a n ~ 2 x 2 + K J ±- / a n ~ s x 3 + 

This result is called the Binomial Theorem. 

400. To prove that it holds for any positive integral expo- 
nent, we multiply hoth members by a + x, thus 

/ x ^i , , n(n—T) , , n(n — l)(n— 2) . „ 
(a+x) n+1 =a n+1 +?ia n x + \ V ^ar^- v J \ ' a n ~ 2 x 3 

, „ n(n— 1) . , 



+ + a n x + na n ~ y x~+ v a n ~ i x 6 +. 

X.A 



= a n + 1 + (n + 1) a n x + 



n (n — 1) 



a"" 1 x- 



ln{n-l)(n-2) ^-1) 1 
+ L 1.2.3 " + 1.2 J +> 

= a n+1 + (w + 1) a n x + -™ 2 In - 1 + 2 1 a n ~ l a? 

+ n S [-H <-*+ 

{ ?i "T~ 1 ) 72- 

= a" +1 + (n + 1) a" a; + ^-j-tj a"- 1 ar 2 



(w + 1) n (n— 1) _„ 3 
1.2.3 ~ a " " * + 



where it is evident that every term except the first will con- 
tain the factor n + 1. 



300 ALGEBRA. 

"We observe that the expansion is of the same form as the 
value of (a + x) n , having n + 1 in the place of n. 

Hence, if the laws of Art. 399 hold for any positive integral 
exponent, n, they also hold when that exponent is increased 
by 1. But we have shown them to bold for (a + cc) 4 , hence 
they hold for (a + x) 5 ; and since they hold for (a + x) 5 , they 
also hold for (a + x) 6 ; and so on. Hence they hold for any 
positive integral exponent. 

401. Since 1.2 = [2, 1. 2. 3 = [3, etc. (Art. 394), the Bino- 
mial Theorem is usually written as follows : 

\ n(n—l) „ o n(n—l)(n—2) _ , „ 
(a+x) n =a n +?i a n ~ l x+ V V ~V+— £ } a n ^x 3 + 

If. £L 

402. If a =1, then, since any power of 1 equals 1, we 
bave 

?i(n — l) , w (n. — 1) (w — 2) „ 

403. In performing examples by the aid of the Binomial 
Theorem, we may use the laws of Art. 399 to find the expo- 
nents and coefficients of the terms. 

1. Expand (a + x) 6 by the Binomial Theorem. 

The number of terms is 7. 

The exponent of a in the first term is 6, and decreases by 1 
in each succeeding term. 

The exponent of x in the second term is 1, and increases by 
1 in each succeeding term. 

The coefficient of the first term is 1 ; of the second term, 6; 
if the coefficient of the second term, 6, be multiplied by 5, the 
exponent of a in that term, and the product, 30, be divided by 
the number of the term, 2, the result, 15, will be the coefficient 
of the third term ; etc. 

Eesult, a 6 +6a 5 x + 15 a 4 x 2 + 20 a 3 x* + 15 a 2 x" + 6 a x 5 + x*. 



BINOMIAL THEOREM. 301 

Note. It will be observed that the coefficients of any two terms taken 
equidistant from the beginning and end of the expansion are the same. 
The reason for this will be obvious if, in Art. 401, x and a be inter- 
changed, which is equivalent to inverting the series in the second member. 
Thus, the coefficients of the latter half of an expansion may bo written 
out from the first half. 

2. Expand (1 + sc) 7 by the Binomial Theorem. " 

Result, l 7 + 7.1 6 . x + 21.1 5 . x 2 + 35.1 4 . x 3 + 35.1 3 . x* + 21.1 s . x 5 
+ 7.1 1 . x° + x 1 ; 

or, 1 + 7 x + 21 x 2 + 35 x 3 + 35 x A + 21 x 5 + 7 x 6 + x\ 

Note. If the first term of the binomial is a numerical quantity, it will 
be found convenient, in applying the laws, to retain the exponents at first 
without reduction, as then the laws for coefficients may be used. The re- 
sult should afterwards be reduced to its simplest form. 

3. Expand (2 a + 3 b)'° by the Binomial Theorem. 

(2a + 3&) 5 =[(2«.) + (36)] 5 

= (2 a) 5 + 5 (2 a) 4 (3 b) + 10 (2 a) 3 (3 bf + 10 (2 a)' 2 (3 b) 3 
+ 5 (2 a) (3 by + (3 b) 5 

= 32 a 5 + 240 a 4 b + 720 a 3 V 2 + 1080 a 2 b 3 + 810 a b* + 243 b 5 , 

Aiis. 

4. Expand (irf 2 — ft -1 ) 6 by the Binomial Theorem. 

( m -i _ n -i)« = [(m _i ) + (- ft- 1 )] 6 

= ( m -^) +6(m"2)5(_ w -i) + i5( m -*) 4 (-«- 1 )' 2 +20(m"^) 3 (-H- 1 ) 3 
+ 15 (m~ty (-ft" 1 ) 4 + G (m~*) (-n- l f+ (-ft- 1 ) 6 ' 

=mr 3 + 6 m~ r% (- ft" 1 ) + 15 m~ 2 (ft- 2 ) + 20 i»~* (- ft" 3 ) 
+ 15 m- 1 (ft- 4 ) + 6 m~- (- ft" 5 ) + (ft" 6 ) 

= m~ 3 — 6 m~* ft- 1 + 15 m~- n~- — 20 m~ - ft~ 3 + 15 m~ x ft" 4 
— 6 m " 2 ft -5 + ft -6 , Ans. 



302 ALGEBRA. 

Note. If either term of the binomial is not a single letter, with unity 
as its coefficient and exponent, or if either term is preceded by a minus 
sign, it will be found convenient to enclose the term, sign and all, in a 
parenthesis, when the usual laws for exponents and coefficients may be 
applied. In reducing, care must be taken to apply the principles of Arts. 
227 and 259. 

EXAMPLES. 

Expand the following by the Binomial Theorem : 



5. 


(1 + o) 5 . 


8. 


(a b — c d)~. 


11. 


(J + <$y. 


6. 


(a + a- 3 ) 6 . 


9. 


O 2 + 3 nr)\ 


12. 


(m~% + 2 ?i s y. 


7. 


(x*-2yy. 


10. 


(a- 2 - 4 x*) b . 


13. 


(a-i — &»aj4)«. 



404. To find the rth or general term of the expansion of 

(a + ,r)\ 

"We have now to determine, from the observed laws of the 
expansion, three things ; the exponent of a in the term, the 
exponent of x in the term, and the coefficient of the term. 

The exponent of x in the second term is 1 ; in the third 
term, 2 ; etc. Hence, in the rth term it will be r — 1. 

In any term the sum of the exponents of a and x is n. 
Hence, in the rth term, the exponent of a will be such a quan- 
tity as when added to r — 1, the exponent of x, will produce 
n ; or, the exponent of a will be n — r + 1. 

The coefficient of the term will be a fraction, of the form 

n(n-l)(»-2) , . , , . , , 

-. — - } — q— ; m which we must determine the last 

X- . w . o 

factors of the numerator and denominator. 

We observe that the last factor of the numerator of any 
tennis 1 more than the exponent of a in that term: hence 
the last factor of the numerator of the rth term will be 
n — r + 2. 

Also, the last factor of the denominator of any term is the 
same as the exponent of x in that term : hence the last factor 
of the denominator of the rth term will be r — 1. 



BINOMIAL THEOREM. 303 

Therefore the 

n (n - 1) (n - 2) (n — r + 2) 

rth term = -^ t^t;-^ — ^^^rr - 1 — ct n ~ r + 1 x r ~\ 

1 . Z . 6 (r — 1) 

1. Find the 8th term of (3 J -2b~ l ) n . 
Here r = 8, n = 11 ; hence, the 

8thtom = 1 l:2°3 9 4 8 5.6 6 7 5 < 3a *> , (- 2t -)' 

= 330 (81 a 2 ) (- 128 b~'') = - 3421440 a 2 J- 7 , ^ras. 

Note. The note to Ex. 4, Art. 403, applies with equal force to examples 
in this article. 



EXAMPLES. 

Find the 

2. 10th term of (a + x) 15 . 5. 5th term of (1 - a 2 ) 12 . 

3. 6th term of (1 + m) u . 6. 9th term of (x~ 1 -2 y*) u . 

4. 8th term of (c - d) l \ 7. 8th term of (a% + 3 x" 1 ) 10 . 

405. A trinomial may he raised to any power by the Bi- 
nomial Theorem, if two of its terms he enclosed in a paren- 
thesis and regarded as a single term ; the operations indicated 
being performed after the expansion by the Theorem has been 
effected. 

1. Expand (2 a — b + c 2 ) 3 by the Binomial Theorem. 

(2 a - b + c 2 ) 3 = [(2 a-b) + (c 2 )] 3 

= (2 a - by + 3 (2 a -b) 2 (c 2 ) + 3 (2 a - b) (c 2 ) 2 + (c 2 ) 3 

=Sa 3 ~12a 2 b + 6ab 2 -b 3 +3c 2 (4:a 2 -4ab + b 2 ) + 3c 4 (2a-b) + c 6 

= 8 a 3 - 12 a 2 b + 6 a b 2 - b 3 + 12 a 2 c 2 - 12 a b c 2 + 3 b 2 c 2 
+ 6ac i -3bc 4 + c 6 , Ans. 



304 ALGEBRA. 

The same method will apply to the expansion of any poly- 
nomial hy the Binomial Theorem. 

EXAMPLES. 

Expand the following by the Binomial Theorem : 

2. (1-x-x 2 )*. 4. (1 - 2 x - 2 x 2 )\ 

3. (x-+3x + l) 3 . 5. (l + a;-a; 2 ) 5 . 



XXXVIII.— UNDETERMINED COEFFICIENTS. 

406. A Series is a succession of terms, so related that each 
may be derived from one or more of the others, in accordance 
with some fixed law. 

The simpler forms of series have already been exhibited in 
the progressions. 

407. A Finite Series is one having a finite number of 
terms. 

408. An Infinite Series is one whose number of terms is 
unlimited. 

The progressions, in general, are examples of finite series ; 
but, in Art. 380, we considered infinite Geometrical series. 

409. An infinite series is said to be convergent when the 
sum of the first n terms cannot numerically exceed some finite 
quantity, however large n may be ; and it is said to be diver- 
gent when the sum of the first n terms can numerically exceed 
any finite quantity by taking n large enough. 

For example, consider the infinite series 

1 + x + x- + X 3 + 

The sum of the first n terms 

1 + x + x 2 + x 3 + +x n ~ 1 = ^^ (Art. 120). 

1 — x 



UNDETERMINED COEFFICIENTS. 305 

If x is less than 1, x n is less than x, however largo n may 
be ; consequently the numerator and denominator of the frac- 
tion are each less than 1, and positive ; and the numerator is 
larger than the denominator ; hence the fraction is equal to 
some finite quantity greater than 1. The series is therefore 
convergent if x is less than 1. 

If x is equal to 1, each term of the series equals 1, conse- 
quently the sum of the first n terms is n ; and this can numer- 
ically exceed any finite quantity by taking n large enough. 
The series is therefore divergent if x = 1. 

If x is greater than 1, each term of the series after the first 
is greater than 1, consequently the sum of the first n terms is 
greater than n ; and this sum can numerically exceed any finite 
quantity by taking n large enough. The series is therefore 
divergent if x is greater than 1. 

410. Every infinite literal series, arranged in order of pow- 
ers of some letter, is convergent for some values of that letter, 
and divergent for other values. 

We will now show that it is convergent when that letter 
equals zero. 

Let the series be 

a + bx + cx 2 + dx s + + Jex n ~ 1 + 

The sum of the first n terms is 

a + bx + cx 2 + dx 3 + + kx n ~\ 

which is equal to a, if x is made equal to 0. 

Hence, however large n may be, the sum of the first n terms 
is equal to a, if x is equal to 0. Consequently the series is 
convergent if x = 0. 

411. Infinite series may be developed by the common pro- 
cesses of Division, as in Art. 101, Exs. 19 and 20, and Extrac- 
tion of Roots, as in Arts. 239 and 243 ; and by other methods 
which it will now be our object to elucidate. 



306 ALGEBRA. 



UNDETERMINED COEFFICIENTS. 

412. A method of expanding algebraic expressions into 
series, simple in its principles, and general in its application, 
is based on the following theorem, known as the 



THEOREM OF "UNDETERMINED COEFFICIENTS. 

413. If the series A + Bx + Cx 2 + Bx 3 + is always 

equal to the series A' + B' x -f- C ar -f- D' x 3 + , for a ny 

value of x which makes both series convergent, the coefficients 
of like powers of ' x in the two series will he equal. 

For, since the equation 
A + Bx + Cx 2 +Dx 3 + =A'+B'x+ C'x-+D>x 3 + 



is satisfied for any value of x which makes both members con- 
vergent ; and since by Art. 410, if x is equal to 0, both mem- 
bers are convergent ; it follows that the equation is satisfied if 
x = 0. Making x — 0, the equation becomes 

A = A'. 

Subtracting A from the first member of the equation, and its 
equal, A', from the second member, we have 

Bx+ Cx 2 +Bx 3 + = B'x + C'x 2 + D'x 3 + 

Dividing through by x, 

B+Cx + Bx 2 + =B'+ C'x + D'x 2 + 



This equation is also satisfied for any value of X which makes 
both members convergent ; hence it is satisfied if x = 0. Mak- 
ing x = 0, we have; 

B = B'. 

Proceeding in this way, we may show C= O, I) = D', etc. 



UNDETERMINED COEFFICIENTS. 307 

Note. The necessity for the limitation of the theorem to values of x 
which make both series convergent, is that a convergent series evidently 
cannot be equal to a divergent series ; and two divergent scries cannot be 
equal, as two quantities which numerically exceed any finite quantity can- 
not be said to be equal. 

Hence, in all applications of the theorem, the results are only true when 
both members are convergent. 



APPLICATION TO THE EXPANSION OF FRACTIONS INTO SERIES. 

2 + 5 x 

414. 1. Expand - — - into a series. 

JL — o x 

We have seen (Art. 101), that any fraction may he expanded 
into a series by dividing the numerator by the denominator ; 
consequently, we know that the proposed development is pos- 
sible. Assume then, 

^ + f x =A+Bx + Cx 2 + Dx*+Ex i + (1) 

1 — ox 

where A, B, C, D, E, are quantities independent of x. 

Clearing of fractions, and collecting together in the second 
member the terms containing like powers of x, we have 



2+5x = A + B 
-3.4 



x+ C 
-3B 



x-+ D 
-3(7 



x 3 + E 
-3D 



x* + 



Equation (1), and also the preceding equation, are evidently 
to be satisfied by all values of x which make the second mem- 
ber a convergent series. Hence, applying the Theorem of Un- 
determined Coefficients to the latter, we have 

A = 2. 
B — 3 A — 5 ; whence, B = 3 A + 5 = 11. 
C-3B = 0; whence, C = 3B =33. 
D - 3 C = ; whence, D = 3 C = 99. 
E-3D = 0; whence, E=3D =297. 



303 



ALGEBRA. 



Substituting these values of A, B, C, D, E, in (1), we 

have 

^+-^ = 2 + 11 a: + 33 z 2 + 99 cc 3 + 297 a: 4 + , 

1 — o x 

which may he readily verified by division. 

This result, in accordance with the last part of the Note to 
Art. 413, only expresses the value of the fraction for such 
values of x as make the second member a convergent series. 

2. Expand — — « into a series. 



l-2ic-a;' 2 



Assume 



1 — 3 x- 



■x* 



A + Bx + Cx 2 +Dx !i + JEx i + 



l—2x—x 

Clearing of fractions, and collecting terms, 
1-Sx 



x 2 = A + B 
-2 A 



x+ C 
-2B 
- A 



x 2 + D 
-2(7 
- B 



x 3 + E 
-2D 

- C 



.v 



Equating the coefficients of like powers of x, 
A = l. 
B-2A = -3; whence, B — 2A — 3 = — 1. 
C-2B-A = -1; whence, C=2B + A-l = -2. 
D-2C—B= 0; whence, D=2 C+B = — 5. 
E-2D- 0— 0; whence, E=2D+ C=-12. 

Substituting these values, 



1-3 



x 



■x 



\-2x 



■X" 



r = 1 — x — 2 x 2 — 5 x 3 — 12 x* — , A ns. 



Note. This method enables us to find the law of the coefficients in any 
expansion. For instance, in Example 1, we obtained the equations C=3B, 
D = BC, E = ZD, etc. ; or, in general, any coefficient, after the second, is 
three times the preceding. In Example 2, we obtained the equations 
Ij — IC-^B, E=2D + C, etc.; or, in general, any coefficient, after the 
third, is twice the preceding plus the next but one preceding. After the 
law of the coefficients of any expansion has been found, we may write out 
the subsequent terms to any desired extent by its aid. 



2-3 


x + Ax 2 


1 + 2 


x — 5x 2 ' 


3 + x 


-2x 2 


3 — x 


+ x 2 ' 


1- 


-3x 2 



UNDETERMINED COEFFICIENTS. 309 

EXAMPLES. 

Expand the following to five terms : 

3. * = ". 6. \~'-* . 9. 

1 + X 1 + x + x z 

3+1* 7 l-^ 2 1Q 

l-5x' l + 2x-3a; 2 ' 

2 - a; + x 2 8 l + 2.x n 
Sl 1-x 2 ' ' 2-cc-a; 2 ' " 2 -3 a: -2 a- 

415. If the lowest power of x in the denominator is higher 
than the lowest power of x in the numerator, the method of the 
preceding article will he found inapplicable. We may, how- 
ever, determine by actual division what will be the exponent 
of x in the first term of the expansion, and assume the fraction 
equal to a series whose first term contains that power of x ; the 
exponents afterwards increasing by unity each term as usual. 

1. Expand ^ ; 2 into a series. 

O X — X 

Proceeding in the usual way, we should assume 
1 



=A+Bx+ 



Clearing of fractions, l = 3Ax + (3B — A)x 2 + 



Equating the coefficients of like powers of x, we have 1 = ; 
a result manifestly absurd, and showing that the usual method 
is inapplicable. 



x- 1 



But, dividing 1 by 3 x — x' 1 , we obtain — — as the first term 

o 

of the quotient ; hence we assume the fraction equal to a series 

whose first term contains x~ l ; next term x°, or 1 ; next term 

x ; etc. Or, 

5 T = Ax~ 1 + B + Cx + Dx 2 + Ex*+ 

OX — X' 



310 ALGEBRA. 

Clearing of fractions, and collecting terms, 

x* + 



1=3A+3B 
- A 



x + 3C 
- B 



x- + oD 

- C 



x 3 + 3E 
- D 



Equating the coefficients of like powers of x, 

3 A = 1 ; whence, A = - . 

o 

3B~A = 0- whence, B = — = - . 

3 9 

7? 1 
3 C—B — 0] whence, C = — = — . 

3D- C=0; whence, D = ^- = — . 

3E-D = 0; whence, E=^ = ~. 

Substituting these values, 

1 1.11 1.1, 



3x-x 2 3 9 27 '81 '243 

EXAMPLES. 

Expand the following to five terms : 

2 n l + z-a 2 . l — 2x* — at 



3sc 2 -2a; 8 " cc-2a 2 +3cc 3 a- a + a: 8 -a- 4 



APPLICATION TO THE EXPANSION OF RADICALS INTO SERIES. 

416. As any root of any expression consisting of two or 
more terms can be obtained by the method of Art. 247, we 
know that the development is possible. 



1. Expand \'l + x 2 into a series by the Theorem of Unde- 
termined Coefficients. 



Assume \Jl-\-x' 2 = A + Bx + Cx*+ Dx* + 22x i + 



UNDETERMINED COEFFICIENTS. 



311 



Squaring both members, we have (Art. 230), 



l + x 2 = A 2 



+ 2AB 



x + B 2 

+ 2 AC 



.'■- 



+ 2 AD 

+ 2BC 



x 3 + 



C 2 
+ 2AE 
+ 2BD 



./■• 



Equating the coefficients of like powers of x, 

A 2 = l; whence, A = l. 





2AB = 0; whence, B = - 

B 2 + 2AC=1; whence, C = 



2 A 

1 - B 2 _ 1 
^2 



2vlZ> + 2i?C Y = 0; whence, D = =0 

C 2 + 2AB+2BD = 0; whence, .E=- 



2 A 

BC 
A 

2BD+ C 2 



2 A 



1 

8 



Substituting these values, 



yi + x 2 = l + -x 2 --;x i + 



2 8 

which may be verified by the method of Art. 239. 

Note. From the equation A- = \, we may have A= ± 1 ; and taking 

-4 = - 1, we should find C = - — , E=— , , so that the expansion might 

2 8 

he as follows : 



2 8 



This agrees with the remark made after the rule in Art. 239. 

EXAMPLES. 

Expand the following to five terms : 

2. Sjl + x. 4. v / l-2a; + 3x 2 . 

3. v/1-2*. 5. y/1 + * + x 2 . 



6. yi-aj. 



7. yl + ic + a; 2 . 



[12 ALGEBRA. 



APPLICATION TO THE DECOMPOSITION OF RATIONAL 

FRACTIONS. 

417. When the denominator of a fraction can be resolved 

into factors, and the numerator is of a lower degree than the 
denominator, the Theorem of Undetermined Coefficients ena- 
bles us to express the given fraction as the sum of two or more 
'partial fractions, whose denominators are the factors of the 
given denominator. 

We shall consider only those cases in which the factors of 
the denominator are all of the first degree. 

CASE I. 

418. When the factors of the denominator are all un- 
equal. 

x + 7 . 

Let — tt— 7z ^r be a fraction, whose denominator is 

(3 x — 1) (5 x + 2) 

composed of two unequal first degree factors. We wish to 

prove that it can be decomposed into two fractions, whose 

denominators are 3 x — 1 and 5 x + 2, and whose numerators 

are independent of x. To prove this, assume 

x + T A B 

+ 



(3 x - 1) (5 x + 2) 3 x - 1 5 x + 2 

We will now show that such values, independent of x, may 
be given to A and B, as will make the above equation identi- 
cal, or true for all values of x. Clearing of fractions, 

x + 7 = A (5 x + 2) + B (3 x - 1) 
or, x + 7 = (5 A + 3 B) x + 2 A - B, 

which is to be true for all values of x. Then, by Art. 413, the 
coefficients of like powers of x in the two members must be 
equal ; or, 

5^+35=1 

2A-B=7 



UNDETERMINED COEFFICIENTS. 313 

From these two equations we obtain A = 2, and B = — 3. 
Hence, the proposed decomposition is possible, and we have 

x + 7 2 -3 



(3x-l)(5x + 2) 3x-l'5x + 2 



2 



o 



3a;-l 5x + 2 

This result may be readily verified by finding the sum of 
the fractions. 

In a similar manner we can prove that any fraction, whose 
denominator is composed of unequal first degree factors, can 
be decomposed into as many fractions as there are factors, 
having these factors for their denominators, and for their nu- 
merators quantities independent of x. 

EXAMPLES. 

1. Decompose — 2 — j- — — — into its partial fractions. 

The factors of the denominator are x — 8 and x — 5 (Art. 118). 

a .i 3 x — 5 A B ,* >. 

Assume, then, — — — = -\ (J-) 

x- — Id x + 40 x — 8 x — 5 

Clearing of fractions, and uniting terms, 

3 x - 5 = A (x - 5) + B (x - 8) 

19 

Putting x = 8, 19 = 3 A, or A = -g-- 

Putting cc = 5, 10= — 3 B, or B= -— g-- 

Note. The student may compare the above method of finding A and 
B with that used on page 312. 

Substituting these values in (1), 

19 10 

3^-5 "3" 3_19 10 

x 1 - 13 x + 40 ~~ x-S + x — 5 ~ 3 (x - 8) ~ 3 (x - 5) ' AnS ' 



314 ALGEBRA. 

EXAMPLES. 

Decompose the following into their partial fractions 

5ic — 2 3a; + 2 n cc 

«• - o — r • ^ i — o — • 

ar — 4 ar — 2 a; 

x + 9 _ 2a; — 3 

0. 



6. 






•> 
ar — 


13 a- + 42 ' 


7. 




17 


6 a; 2 


- 13 x - 5 




a; 


* 



x 1 + 3 a; ' a; 2 — 3 a; — 4 

ft 7a ' + 9 Q 

9 + 9 x - 4 a; 2 * * (a; 2 - 1) (x-2)' 



CASE II. 
419. When the factors of the denominator are all equal. 

x 2 — 11 x + 26 

1. Separate '- 3 — into its partial fractions. 

\x — o) 

If we attempt to perform the example hy the method of 
Case I, we should assume 

x*—l±x + 2& ABC 

+ o + 



(x — 3) 3 x — 3 x — 3 x — 3 
This would evidently he impossible, as the sum of the frac- 
tions in the second member is — ; which, as A, B, 

x — o 

and C are, by supposition, independent of x, cannot be equal 
a; 2 -11 a; + 26 

tO ; jr-r . 

(x — 3) 8 

The method to be used in Case II depends on the following : 

n ., .,. a x"- 1 + b x n ~ 2 + c x"-* + + k 

Consider the traction r . 

(aj + h) n 

Putting x = y — h, the fraction becomes 
a(y- hy- 1 + b(i/ — h) n - 2 + e (y — A)"" 3 + + k 



UNDETERMINED COEFFICIENTS. 315 

If the terms of the numerator are expanded by the binomial 
theorem, and the terms containing like powers of y collected 
together, we shall have a fraction of the form 

a, y' 1 ' 1 + b r y n ~ 2 + c x y n ~ 3 + +ky 



y , 



Dividing each term of the numerator by y n , we have 
a x b x c x l\ 

y y y v 

Changing back y to x + h, this becomes 

«i b, c, /.', 

+ , , ' , + , ,' Q + + 



x + h (x+ h) 2 (x + h) 3 (x + h) n 

This shows that the assumed fraction can be expressed as 
the sum of n partial fractions, whose numerators are indepen- 
dent of x, and whose denominators are the powers of x + h, 
beginning with the first, and ending with the nth. 

In accordance with this, we assume 

a 2 -11 a; + 26 A B , C 



( x -3) 3 ~ x - 3 T (x - 3) 2 ^ (x - 3) 3 ' 
Clearing of fractions, 

x 2 -llx + 26 = A(x-3)- + B(x-3)+ C 

= A (x 2 -6x + 9) + B(x — B)+-C 
= A x 2 + (B - 6 A) x + 9 A - 3 B + C. 

Equating the coefficients of like powers of x, 

A = l, B-6A = -11, and 9A-3B+ C=26 
Whence, A = 1, B = -5, and C = 2. 

Substituting these values, 

a; 2 -lla; + 26_ 1 5 2 

(x - 3) 3 ~~ " ~x~^3 ~ (x - 3) 2 + (x - 3) 3 ' US ' 



2' 



316 ALGEBEA. 

EXAMPLES, 

Separate the following into their partial fractions : 

ar+3a: + 3 a; 2 3 a; — 10 

(x + 1) 3 " ' (x-2) 3 * (2a--5) s 

2 a; -13 3 a; 2 -4 18 a; 2 + 12 a; - 3 

(x-5) 2 ' (cc + 1) 8 ' (3 x + 2/ 



CASE III. 

420. When some of the factors of the denominator are 
equal. 

1. Separate zrrz into its partial fractions. 

1 x (x + l) 3 l 

The method in this case is a combination of the methods of 
Cases I and II. We assume 

3 a + 2 A B C D 

+ ,.. . ,,, + ,.. , ^„ + — • 



a; (x + l) 3 a- + 1 (x + l) 2 (a; + l) 3 
Clearing of fractions, 
3ai + 2 = A x (x + l) 2 + B x {x + 1) + Cx + D (x + l) 3 

= (A + D) x* + (2A + B + 3 D) x 2 + ( A +B + C+ 3 D) x +B. 

Equating the coefficients of like powers of x, 

£> = 2, A + B + C+3 B = 3, 2A + B+3B = 0, and 

A + B = 

Whence, A = -2, B = -2, C = l, andZ> = 2. 
Substituting these values, 

3a; + 2 2 2 12, 



a;(a; + l) 3 a; + 1 (a? + 1) 2 (a- + l) 3 x 

It is impossible to give an example to illustrate every pos- 
sible case; but no difficulty will be found in assuming the 



UNDETERMINED COEFFICIENTS. 317 

proper partial fractions, if attention be given to the following 
general case. A fraction of the form 

X 

(x + a) (x + b) (x + m) r (x + n) s 

should he put equal to 

A B E t F K 

x+a x+b x + m (x+m)' (x + m) r 

L M R 

+ x + n + (x + n) 2 + + (x + n) s + 

Single factors, like x + a and x + b, having single fractions 

A B 

like and - — — , corresponding ; and repeated factors, 

x+a -x+b 

like (x + m) r , having r partial fractions corresponding, ar- 
ranged as in Case II. 

EXAMPLES. 

Separate the following into their partial fractions : 

8-3 x-x' 2 15 - 7 x + 3 x 2 - 3 x z 

x(x + 2) 2 ' x 3 (x + 5) 

3 ,-r 3 - 11 x 2 + 13 x - 4 6 ar - 14 x + 6 

x (x - 1) (x - 2) 2 ' (x-2)(2x-3y 2 ' 

3a;-l ? 5 x 2 + 3 x + 2 

x 2 {x + iy ' x 3 (x + iy 2 ' 

421. Unless the numerator is of a lower degree than the 
denominator, the preceding methods are inapplicable. 

:' .'•- + 1 . 

For example, let it be required to separate -— ^— — into its 
partial fractions. Proceeding in the usual way, we assume 
2 .r- +1 A B 

Ob *jC JC *fc -x. 

Clearing of fractions, 

2 x 2 + 1 = A (x - 1) + B x = (A + B) x - A. 



318 



ALGEBRA. 



Equating the coefficients of like powers of x, we have 2 = 0; 
an absurd result, and showing that the usual method is inap- 
plicable. 

But by actual division, as in Art. 150, we have 



2z 2 +l 2« + l 

— £ + 



x- — x 



x- 



X 



2 x + 1 . .,„.-, 

We may now separate — ^ into its partial fractions by 

the usual method, obtaining 

2a; + 1 1 3 



x' — x 



X X 



1" 



Hence, 



2a 2 +l 2^ + 1 1 3 

= 2 + —r- =2- - + - —,Ans. 



x- 



X 



x- — X 






APPLICATION TO THE REVERSION OF SERIES. 

422. 1. Given y = 2x + x 2 — 2x z — 3x* + , to revert 

the series, or to express x in terms of y. 

Assume x = Ay + B if + G if + D if + (1) 

Substituting in this the given value of y, we have 

x=A(2x + x 2 -2x s -3x i +...)+B(ix 2 + x 4 +Ax s -Sx i + ...) 
+ <7(8x 3 + 12z 4 +...) + Z>(16z 4 +...) + 



or, 



X: 



2Ax+ A 


x 2 -2A 


x 3 - 3 A 


+ 4:B 


+ 4B 


- IB 




+ 8 


+ 12 G 
+ 16Z> 



x* + 



Equating the coefficients of like powers of x, 



2 A = 1 ; whence, A = - . 



A + 4 B— ; whence, B— r = — - , 

4 8 



-2A + 4.B+S G=0; whence, 0= 



3^ 
16" 



-3^-7^+12 C+1GZ> = 0; whence, D- — 



13 
128' 



UNDETERMINED COEFFICIENTS. 



319 



Substituting these values in (1), 

V if 3 y 3 13 y i 

If the even powers of x are wanting in the given series, we 
may abridge the operation by assuming x equal to a series 
containing only the odd powers of y. 

Thus, to revert the series y = x — x 3 + x 5 — x 1 + , we 

assume x = A y + B y 3 + C y 5 + D y" + 

If the odd powers of x are wanting in the given series, the 
reversion of the series is impossible by the method previously 
given. But by substituting another letter, say t, for x~, we 
may revert the series and obtain a value of t, or of x 2 , in terms 
of y ; and by taking the square root of the result, express x 
itself in terms of y. 

If the first term of the series is independent of x, we cannot, 
by the method previously given, express x definitely in terms 
of y ; though we can express it in the form of a series in which 
y is the only unknown quantity. 

2. Kevert the series y = 2 + 2x — x 2 — x 3 + 2x i + 

We may write the series, 

y-2 = 2x-x 2 -x 3 + 2x 4 + (1) 

Assume x=A(y-2) + B(y-2y 2 +C(y-2) 3 +D(y-2y+... (2) 
Substituting in this the value of y — 2 given in (1), we have 

x = A(2x— x 2 — x s +2x*+ ...) + B(4:X 2 + x i — ix 3 — 4x*+ ...) 

+ C(Sx 3 -12x i + ...) + D(lGx 4 + ...)+ 



or, x = 2 A x — A 

+ 4:B 



x 2 - A 


x 3 + 2 A 


-42? 


- 3B 


+ SC 


-12C 




+ 16B 



x* + 



320 ALGEBRA. 

Equating the coefficients of like powers of x, 

2 A — l; whence, A = ^ . 

— A + 4 B =:Jd ; whence, B = ^ . 

o 

— A-4:B + 8C = 0; whence, C = ^. 

o 

7 
2A-3B-12C+16I> = 0] whence, D = -j-. 

Suhstituting in (2), 

x=l(y-2)+l( 1/ -2y+l( }/ -2y+^( u -2y+ ,Am. 

EXAMPLES. 

Revert the following series to four terms : 

3. y = x + x- + x 3 + x i + 

4. ij = 2x + 3x 3 + 4:x 5 + 5x 7 + 

5. i/ = x — x 3 + x 5 — x 1 + 

2 3 4 

^y»— rj/*" 'V*^ 

\K/ *Aj *AJ 



tAs \Kj \Aj 



8. y = 3x-2x 2 +3x 3 -4x i + 



Note. This method may sometimes be used to find, approximately, the 
root of an equation of higher degree than the second. Thus, to solve the 
equation 

we may put .1 = 2/, and revert the series ; giving, as in Ex. 1, Art. 422, 

1 1 „ 3 , 13 

x = — ii if-\ if if + 

2 & J 16 128 

Putting back y=.l, we have 

.1 .01 .003 .0013 
^ 2 8 16 128 
= . 05-. 00125 + . 00019 -. 00001 + = .04893 + , Ans. 

This method can, of course, only be used when the series in the second 
member is convergent. 



BINOMIAL THEOREM. 321 

XXXIX. — BINOMIAL THEOREM. 

ANY EXPONENT. 
423. We have seen (Art. 402) that when n is a positive 



integer, 



n(n— 1) „ n(n — l)(n — 2) „ 
(l + a; )»=l + wa; + — ^r V+ — ^ '- x z + 



We shall now prove that this formula is true when n is a 
positive fraction, a negative integer, or a negative fraction. 

1. Let n be a positive fraction, which we will denote by 

p ... 

— ; p and q being positive integers. 



Now (Art. 252), (1 + x) * = V (1 + xf 



= yi+px+ , (Art. 402). 

Extracting the ^th root of this expression by the method of 



Art. 247, 



1 +px + 
1« = 1 



. p x 

1 + — + 



q j) x 



That is, (l + a:)i" = l + — + (1) 

2. Let n be a negative quantity, either integer or fraction, 
which we will denote by — s. 

Then (Art. 255), 

= , (by Arts. 402, and 423, 1). 

1 + sx+ ' v J 

From which, by actual division, we have 

(l + x)-° = l-sx + (2) 



322 ALGEBRA. 

From (1), (2), and Art. 402, we observe that whether n is 
positive or negative, integral or fractional, the form of the 
expansion is 

(1 + x) n = 1 + n x + A x 2 + B x 3 + Cx* + (3) 

x 
Writing - in place of x, we have 
a 

/y* /y>-J /y.3 /y»4 



1 + -) =l + n- + A~ r2 + B-+C— i + 

al a a 1 a 6 a 4 

Multiplying this through by a n , and remembering that 

(x\ n r / x\ i " 
1+-) = a\l+-j = (a + x) n , we have 

(a + x) n = a n + n a"' 1 x + A a n ~ 2 x- + Ba n ~ s x s + (4) 

To find the values of A, B, etc., we put x + z for x in (3), 
and regarding (x + z) as one term, we shall have 

[1 + (x + z)'] r ' = l + ii(x + z) + A(x + z) 2 + B(x + z) 3 + 

= 1 + n x + A x' 1 + B x s + 

+ (n + 2Ax + 3Bx 2 + )*+ (5) 

Regarding (1 + x) as one term, we shall have, by (4), 

[(1 + x) +z~] n = (1 + x) n + n (l + x) n - 1 z+ (6) 

Since [1 + (x + z)~\ n = [(1 + x) + z\ n , identically, we have 
from (5) and (6), 

l + nx + Ax 2 +Bx z + + (n + 2Ax + 3Bx 2 + )z+ 

= (1 + a-) n + w(l + a-) n_1 * + 

which is true for all values of z which make both members of 
the equation convergent. Hence, by Art. 413, the coefficients 
of z in the two series must be equal ; or, 

11 (l + x) n - 1 = n + 2Ax + 3Bx 2 + 



BINOMIAL THEOREM. 323 

Multiplying both members by 1 + x, 

n(l + x) n = n+ (2A + n)x+(3B+2A)x 2 + 

or, by (3), 

n + n 2 x + n Ax 2 + n Bx 3 + = n+ (2 A + n) x 

+ (3B+2A)x 2 + 

•which is true for all values of x which make both members of 
the equation convergent ; hence, equating the coefficients of 
like powers of x, 

sfy {fl, j\ 

2 A + n = n 2 ; whence, 2 A = n 2 — n, or A = - --= 

3B+2A = ?iA; whence, 3 B=n A-2 A = A (n-2) 

„ A (n - 2) n (n - 1) (n - 2) 
B =~3— = \3 

Substituting in (4), 

(a + x) n = a n + na"- 1 x+ V ^\~ ' a n ~ 2 x 2 



n(n-l)(n-2) 



+ — i -^ a n ~ a X s + 



which has tbus been proved to hold for all values of n, positive 

or negative, integral or fractional. Hence, the Binomial The- 
orem has been proved in its most general form. The result, 
however, only expresses the value of {a + x) n for such values 
of x as make the second member convergent (Art. 413). 

424. When n is a positive integer, the number of terms in 
the expansion is n + 1 (Art. 399). When n is a fraction or 
negative quantity, the expansion never terminates, as no one 
of the quantities n — 1, n — 2, etc., can become equal to zero. 
The development in that case furnishes an infinite series. 



324 ALGEBRA. 

425. The method and notes of Art. 403 apply to the ex- 
pansion of expressions hy the Binomial Theorem when the 
exponent is a fractional or negative quantity. 

2 

1. Expand (a + x) :! to five terms. 

2 
The exponent of a in the first term of the expansion is - , and 

o 

decreases hy one in each succeeding term. 

The exponent of x in the second term of the expansion is 1, 
and increases hy one in each succeeding term. 

The coefficient of the first term is 1 ; of the second term, 

2 2 

- ; multiplying the coefficient of the second term, -, hy the ex- 

o o 

1 2 

ponent of a m that term, — - , and dividing the product, — - , 

hy the numher of the term, 2, we ohtain — - as the coefficient of 
the third term ; etc. 

2 2_i 1 _4 4 -i 7 _jo 
Kesult, a 3 + ~a 3 x—-a ^x 2 +^ra 3 x 3 — 7 r-r^a s x A -\- 



.^1-2 



2. Expand (1 + 2 a: 2 ) -2 to five terms. 

(l + 2zV 2 =D- + (2*-)]- 

= l- 2 - 2.1-= . (2 x$) + 3.1- 4 . (2 xfy - 4.1- 5 . (2 a;*) 8 

+ 5.1- 6 . (2 a;*) 4 



i 



= 1 - 2 (2 x-) + 3 (4 a) - 4 (8 x?) + 5 (16 a 2 ) - .... 
= l-4a 2 + 12x-32a:- + S0a; 2 — , Ans. 

3. Expand (a -1 — 3 a: *) * to five terms. 

(a- 1 -3a 5 ~ i )~* = [(«" 1 ) + (-3 a; - *)]"* 



BINOMIAL THEOREM. 325 

= (a- 1 )"^ - \ (co- 1 )-'" (- 3 aT*) + " (0~ V (- 3 x~fy 

1 40 L3 _ JL 455 —JUS 1 

-TFr(0~ v (-3* l ) 8 +iS(«- 1 ) ¥ (-3* *)«- 



81 v y v y ' 243 

447 _i 14 lo 140 13 _a 

= ft 3_* :w 3a; 2 ) + -3 (9 a; -i)__^ a V(_27 a; 2) 

O 9 ol 

+i-*>o- 

4 7 _i J-" 140 13 _a 455 .lb 

= a^+4a 3 a; 2 + 14a 3 cc _1 + — — a 3 x 2 + -— - a 3 #~ 2 , 

o o 

^l?zs. 

EXAMPLES. 

Expand the following to live terms : 

4. O + z) 2 8. . 8/r — — ' 12. (m~"3-_2« 2 ) -2 . 

v y yl + a; v ' 

5. (1 + ^)- 6 . 9. ^^y 3 - 13. (l + e.r 1 )^. 

6. (l-x)~%. 10. — — . 14. (a; 4 + 4aJ)t 

c 2 + d 

7. V^ 1 ^- 11. (aT*-8y)*. 15- (g -i_ 8y -y 

426. The expression for the rth term, derived in Art. 404, 
holds for any value of n, as it was deduced from the expansion 
which has been proved to hold universally. 

1. Find the 7th term of (1 — cc)~ 3 . 

Here r = 7, n = — ^ ; hence, the 
o 

1 4 _7 10 13 16 
~3 ,_ 3*~3'""3"' ' "3"" " 3 . NR 728a; 6 
j th term = 17 2.3.4.5.6 ( ~ x) = "656T ' 

Ans. 



326 ALGEBRA. 

2. Find the 8th term of (eft + x~%)~ z . 

Here r = 8, n = — 3 ; hence, the 

-3.-4.-5.-6. — 7.-8.-9 , ^ „ . _* , 
8thterm = 0.3.4.5.6.7 ^- W (* ^ 

= — 36 ar b x 3 , Ans. 

EXAMPLES. 

Find the 

3. 8th term of \/ a + x. 7. 7th term of (x' 1 — ifift. 

4. 7th term of (1 + m)- 4 . 8. 5th term of . 

(n~* — <ry 

5. 5th term of (1 — ar)~ *. 9. 6th term of (eft + 3 x~ l )~%. 

1 -2 

6. 6th term of . 10. 8th term of (x 3 y — z 3 ) -3 . 

y'x 2 + f 

427. To find any root of a number approximately by the 
Binomial Theorem. 

1. Find the approximate square root of 10. 

y/ 10 = 10* = (9 + 1) * = (3 2 + 1)* 
Expanding this hy the Binomial Theorem, 

(3' + 1)* = (3*)* + \ (3 a )-i - \ (3')-* + 1 (3T* 
111 ^ 

— 3 _L Z 3-1 _ Z Q-8 i Q-5 ^ q-7 i 

-6 + 2 .6 8 .3 +jg. 3 -J28-3 + 

= 3-i _1_ J_ _t_ _JL 

+ 2.3 8.3 3 + 16.3 6 128. 3 7+ 

= 3 + .16667 - .00163 + .00026 - .00002 + 

= 3.16228+, 



BINOMIAL THEOREM. 327 

which is the approximate square root of 10 to the fifth decimal 
place, as may be verified by evolution. 

2. Find the approximate cube root of 26. 

$ 26 = 26* = (27 -1)3 = (3 3 - 1)* 
Expanding this by the Binomial Theorem, 

(3 3 - 1)* = (3 3 )^ + \ (3 3 )- 3 * (- 1) - i (3 3 )"* (- 1)» 

+| [ (3 3 r l (-i) 3 - 



1 1 K 
Q Q-2 Q-5 Q-8 

-°~r 6 ~9 "si" 5 ~ 



= 3 



3.3 2 9.3 5 81. 3 8 

= 3 - .037037 - .000457 - .000009 - 
= 2.962497 + , Ans. 



RULE. 

Separate the given number into two parts, the first of whir// 
is the nearest perfect power of the same degree as the required 
root. Expand the result by the Binomial Theorem. 

Note. If the second term of the binomial is small, the terms in the 
expansion converge rapidly, and we obtain an approximate value of the 
required root by taking the sum of a few terms of the development. But 
if the second term is large, the terms converge slowly, and it requires the 
sum of many terms to insure a considerable degree of accuracy. 

EXAMPLES. 

Find the approximate values of the following to five deci- 
mal places : 

3. #31. 5. #99. 7. #17. 

4. #9. 6. #29. 8. #78. 



328 ALGEBRA. 



XL. — SUMMATION OF INFINITE SERIES. 

428. The Summation of a Series is the process of finding 
a finite expression equivalent to the series. 

Different series require different methods of summation, 
according to the nature of the series, or the law of its forma- 
tion. Methods of summing arithmetical and geometrical series 
have already been given (Arts. 369, 377, and 380). Methods 
applicable to other series will now be treated. 

RECURRING SERIES. 

429. A Recurring Series is one in which each term, after 
some fixed term, bears a uniform relation to a fixed number of 
the preceding terms. Thus 

l + 2x + 3x 2 + Ax 3 + 

is a recurring series, in which each term, after the second, is 
equal to the product of the preceding term by 2 x, plus the 
product of the next term but one preceding by — x 2 . 

The sum of these constant multipliers is called the scale of 
relation of the series, and their coefficients constitute the scale 
of relation of the coefficients of the series. For example, in 

the series 1 + 2 x + 3 x' 2 + 4 x s + , the scale of relation is 

2 x — x 2 , and the scale of relation of the coefficients is 2 — 1. 

430. A recurring series is said to be of the first order 
when each term, commencing with the second, depends on the 
one immediately preceding; of the second order, when each 
term, commencing with the third, depends ujion the tiro im- 
mediately preceding ; and so on. 

If the series is of the first order, the scale of relation will 
consist of one term ; if of the second order, it will consist of 
two terms ; and, in general, the order and the number of terms 
in the scale of relation will correspond. 

431. To find the scale of relation of the coefficients of a 
recurring series. 



SUMMATION OF INFINITE SERIES. 329 

1. If the series is of the first order, it is a simple geometri- 
cal progression, and the scale of relation of the coefficients is 
found by dividing the coefficient of any term by the coefficient 
of the preceding term. 

2. If the series is of the second order, let a, b, c, d, 

represent the consecutive coefficients of the series, and p + q 
their scale of relation. Then, 



c =p b + q a) 
d=p c + q b j 



(4> 



to determine p and q ; solving, we obtain 

ad — bc .. c 2 — b d 

p = 77 , and q = — — . 

ac — b 1 ac — b 1 

3. If the series is of the third order, let a, b, c, d, e, f, 

represent the consecutive coefficients of the series, and p + q 
+ r their scale of relation. Then, 

d =p c + q b + r a 

e =p d + q c + r b 

f=X> e + q d + r c 

from which we can find p, q, and ;■. 

432. To ascertain the order of a series, we may first make 
trial of a scale of two terms, and if the result does not corre- 
spond with the series, we may try three terms, four terms, and 
so on, till the true scale of relation is found. If we assume 
the series to be of too high an order, the terms of the scale 

will take the form r. • 

433. To find the sum of a recurring series, when the scale 
of relation of its coefficients is known. 

Let 

a + bx + cx 2 + dx 3 + +jx n ~ 3 + kx n ~ 2 + lx n ~ l + 

be a recurring se"ries of the second order. Let S denote the 



330 ALGEBKA. 

sum of n terms of the series ; and let p + q be the scale of re- 
lation of the coefficients. Then, 

S= a + b x + c x 2 + d x 3 + + lx n ~ x 

p Sx=pax+pbx 2 +pcx 3 + + pkx n ~ 1 +plx n 

q Sx 2 = q ax 2 + q b X s + + qjx n ~ 1 +q k x n + qlx n + 1 

Subtracting the last two equations from the first, 

S—p Sx — q Sx 2 = a + bx — pax— plx n — q kx n — qlx n + 1 

the rest of the terms of the second member disappearing, be- 
cause, since p + q is the scale of relation of the coefficients, 

c =p b + q a, d =p c + q b, I =p k + qj. 

Therefore we have 

a + (b—p a) x — (p I + q k) x n — q I x n + 1 



S = 



■px — q x" 



the formula for finding the sum of n terms of a recurring series 
of the second order. 

But if n becomes indefinitely great, and the series is con- 
vergent, then the limiting values of the terms which involve 
x n and x n + ! must become 0, and we have at the limit 

s= a + (p-pa)x 
1 — p x — qx 2 

the formula for finding the sum of an infinite recurring series 
of the second order. 

If q = 0, then the series is of the first order, and conse* 
quently b =p a; then, 

1 —px 

the formula for finding the stun of an infinite recurring series 
of the first order. (Compare Art. 380.) 



SUMMATION OF INFINITE SERIES. 331 

In like manner, we should obtain 

a + (b — p a) x + (c —pb — qa)x 2 

o — -z o 5 (o) 

1 —p x — q x- — r x 6 w 

the formula for the summation of an infinite recurring series 
of the third order. 

434. A recurring series, like other infinite series, originates 
from an irreducible fraction, called the generating fraction. 
The summation of the series, therefore, reproduces the frac- 
tion ; the operation being, in fact, the exact reverse of that in 
Art. 414. 

435. 1. Find the sum of l + 2cc + 8ar+28z 3 +100x 4 + 

We must first determine the scale of relation of the coeffi- 
cients. In accordance with Art. 432, we first assume the 
series to be of the second order. We have a = i, b = 2, c==8, 
d = 28. Substituting in the values of p and q derived from 
(^4), Art. 431, we have p = 3 and q = 2. To ascertain if this 
is the proper scale of relation, consider the fifth term, 100 x 4 ; 
this should be 3 x times the preceding term, plus 2 x 2 times 
the next preceding term but one, or, 84 x 4 + 16 x*. This shows 
that the series is of the second order. 

Substituting in (1) the values of a, b, p, and q, we have 

l + (2 — 3)a; _ 1 — x 
b ~ i-Zx-2tf ~ l-3x-2x 2 ' 

EXAMPLES. 

Find the sum of the following series : 

2. l + 2a- + 3x 2 + 5a; 3 + 8a; 4 + 

a ac a c 2 „ a c 3 

3> b^¥ x + !F x "-l^ x+ 

4. 4 + 9a; + 21ar+51;r 5 + 

5. l + 3x + 5x 2 + 7x 3 + 



332 ALGEBRA. 

6. 2-a + 2a 2 -5a 3 +10a i -17a 5 + 

7. 3 + 5x + 7 x' 2 + 13.x' 3 + 23 a; 4 + 45 x* + 

8. l + 3x + 4:x 2 +7x 3 +llx 4 + 

9. 2 + 4x-x*-3x s + 2x i + ±x 5 + 

DIFFERENTIAL METHOD. 

436. The Differential Method is the process of finding any 
term, or the sum of any number of terms, of a regular series, 
by means of the successive differences of its terms. 

437. If, in any series, we take the first term from the sec-, 
ond, the second from the third, the third from the fourth, and 
so on, the remainders will form a new series called the first 
order of differences. 

If the differences be taken in this new series in like manner, 
we obtain a series called the second order of differences ; and 
so on. 

Thus, if the given series is 

1, 8, 27, 64, 125, 216, 

the successive orders of differences will be as follows : 

1st order, 7, 19, 37, 61, 91, 

2d order, 12, 18, 24, 30, 

3d order, 6, 6, 6, 

4th order, 0, 0, 

Hence, in this case there are only three orders of differences. 

438. To find any term of a series. 
Let the series be 

a l) a 2> a 3> a i) a 5> a n1 a n+l> 

Then the first order of differences will be 



SUMMATION OF INFINITE SERIES. 333 

the second order of differences will be 

a 3 — 2a 2 + a l , a A — 2a 3 + a 2 , a 5 — 2a i + a 3 , , 

the third order of differences will be 

a A — 3 a 3 + 3 a 2 — a x , a 5 — 3 a A + 3 a 3 — a 2 , , 

the fourth order of differences will be 

a 5 — 4 a A + 6 a& — 4 a 2 + a 1} , 

and so on ; where each difference, although a compound quan- 
tity, is called a term. 

Let now d x , d 2 , d s , d A , represent the first terms of the 

several orders of differences. Then, 

d l = a 2 — a l ; whence, a 2 = a l + d A . 

d 2 = a 3 — 2 a 2 + a x ; whence, a 3 — 2a 2 — a l +d 2 ^2ai + 2d 1 

— a x + d 2 = a A + 2 d A + d 2 . 

d 3 = a A — 3 a s + 3 a 2 — a x ; whence, a A = a x + 3 d v + 3 d 2 -f- d s . 

d A = a 5 — A a A + 6 a 3 — Aa 2 +a l ; whence, a 5 = a l + 4^d 1 + 6d 2 
+ ±d 3 +d A . 

We observe that the coefficients of the value of a 2 are the 
same as the coefficients of the first power of a binomial; the 
coefficients of the value of a 3 are the same as the coefficients 
of the second power of a binomial ; and so on. Assume that 
this law holds for the nth term ; that is, that the coefficients 
of the value of a H are the same as the coefficients of the (n — l)th 
power of a binomial ; then, 

, „ . (n-l)(n-2) . 

a* = «i + (n - 1) (h + Q d 2 

+ („-!)(„- 2) <«-S) d3+ (1) 

If the law holds for the nth term in the given series, it will 
also hold for the «th term in the first order of differences ; or, 



334 ALGEBRA. 

a n+1 -a n = d 1 +(n-l)d 2 + ^—-^f—^d ;i + (2) 



Adding (1) and (2), we have 

(n - 1) (n - 2) 



a 



, +1 = «!+[! + (n-l)]d l + 



(n - 1) + 



+ 



(w-l)(w-2) (w-l)( w -2)(ra-3) 

~|2- ~\T 



2 



,/., 



= aj + n dx H r^- [2 + n — 2] d 2 

(n - 1) (« - 2) ro 



, n(n-l) ? »(»-l)(n-2) . ,„ 

= «! + »!(/!+ -^ ~ <4 + ^ "^3 + (3) 



where the coefficients are the same as the coefficients of the rath 
power of a binomial. Hence, if the law holds for the nth term, 
it also holds for the {n + l)th term ; but we have shown it to 
hold for the fifth term, a 5 ; hence it holds for the sixth term ; 
and so on. That is, Formula (1) holds for any term in the 
series. 

When the differences finally become 0, the value of the nth 
term can be obtained exactly ; but, in other cases, the result is 
merely an approximate value. 

439. To find the stem of any number of terms of a series. 

Let the series be 

a, b, c, d, e, (1) 

Let S denote the sum of the first n terms. Assume the 

series 

0, a, a + b, a + b + c, a + b + c + d, (2) 

in which the (n + l)th term is obviously equal to the sum of n 
terms of the given scries ; that is, S is the (n + l)th term of 
series (2). Now the first order of differences of series (2) is 



SUMMATION OF INFINITE SERIES. 335 

the same as series (1) ; hence, the second order of differences 
of series (2) is the same as the first order of (1) ; the third 
order of (2) is the same as the second order of (1) ; and so on. 

Then, letting a', d\, d' 2 , d' s , represent the first term, and 

the first terms of the several orders of differences of (2), we 

have a' = 0, d\ = a, d' 2 = d l ,d' s = d 2 , where a, d x , d 2 , 

are the first term, and the first terms of the several orders of 
differences of (1). But, by (3), Art. 438, the (n + l)th term 
of series (2) will be 

n (n - 1) „ n (n - 1) (n - 2) 

In this put for a', d\, d' 2 , d' 3 , their values; then 

n (n — 1) , n (n — 1) (re — 2) . 

S=na + — ^|2 — L d 1 +- ~" L d,+ (3) 

440. 1. Find the 12th term of the series 2, 6, 12, 20, 

30, 

The successive orders of differences will be as follows : 

1st order, 4, 6, 8, 10, 

2d order, 2, 2, 2, 

3d order, 0, 0, 

Then a x = 2, d Y = 4, d 2 = 2, d s , d A , = 0, and n = 12. 

Substituting in (1), Art. 438, the 12th term 

(12 — 1) (12 — 2) 
= 2 + (12-l)4+ l >) } ; 2 = 2 + 44 + 110=156,^. 

2. Find the sum of 8 terms of the series 2, 5, 10, 17, 

1st order of differences, 3, 5, 7, 

2d order of differences, 2, 2, 

3d order of differences, 0, 

Then a = 2, d, = 3, d, = 2, n = S. 



336 ALGEBRA. 

Substituting these values in (3), Art. 439, we have 

= 16 + 84 + 112 = 212, Ans. 

EXAMPLES. 

3. Find the first term of the fifth order of differences of 
the series 6, 9, 17, 35, 63, 99, 

4. Find the first term of the sixth order of differences of 
the series 3, 6, 11, 17, 24, 36, 50, 72, 

5. Find the seventh term of the series 3, 5, 8, 12, 17, 

6. Sum the first twelve terms of the series 1, 4, 10, 20, 
35, 

7. Sum the first hundred terms of the series 1, 2, 3, 4, 
5, 

8. Find the 15th term of the series l 2 , 2 2 , 3 2 , 4 2 , 

9. Sum the first n terms of the series l 3 , 2 3 , 3 3 , 4 3 , 5 3 , 

10. Sum the first n terms of the series 1, 2 4 , 3 4 , 4 4 , 5 4 , 6 4 , 



11. If shot be piled in the shape of a pyramid, with a trian- 
gular base, each side of which exhibits 9 shot, find the number 
contained in the pile. 

12. If shot be piled in the shape of a pyramid, with a square 
base, each side of which exhibits 25 shot, find the number 
contained in the pile. 

INTERPOLATION. 

441. Interpolation is the process of introducing between 
terms of a series other terms conforming to the law of the 

series. 



SUMMATION OF INFINITE SERIES. 337 

Its usual application is in finding intermediate numbers 
between tbose given in Mathematical Tables, which may be 
regarded as a series of equidistant terms. 

442. The interpolation of any intermediate term in a 
series, is essentially finding the nth term of the series, by the 
differential method (Art. 438). Thus, 

Let t represent tbe term to be interpolated in a series of 
equidistant terms, and p the distance the term t is removed 
from the first term, a, expressed in intervals and fractions of 
an interval ; that is, p being the distance to the nth term, 
p = n — 1 intervals. 

In Formula (1), Art. 438, putting p for n — 1, the nth. term 

t=a+pdl+ pj£ ! =v. il+ p(*- 1 np-*> dt+ 

443. 1. In the series :j^ , 7-7 > TE > T£ > T?? > , find the 

13 14 15 16 17 

middle term between 5-= and ^-7 . 

15 lb 

Here, the first differences of the denominators are 

1, 1, 1, 1, 

The second differences are 

0, 0, 0, 

Whence, d t = 1, and d 2 = 0. 

5 
The distance to the required term is 2\ intervals, or p = ^- 

Make a = 13, the denominator of the first term ; then by the 
preceding formula, the denominator of the required term, 

1 2 

Therefore the required term is — or 757, Ans. 

31 ol 

~2 



338 ALGEBRA. 

2. Given ^94 = 9.69536, ^95 = 9.74679, y/ 96 = 9.79796 ; 
to find y/94i. 

Here, the first differences are 

.05143, .05117, 

and the second differences are 

-.00026, 

Whence, ^ = .05143, d 2 = -. 00026, 



. 1 . 1 

The distance of the required term is - interval, or p = -? , 

Then the required term, 

1(|-1 
* = 9.69536 + 1 x .05143 + 4 ^ (- ,0( 

= 9.69536 + .01286 - J, (- .00026) + 

= 9.69536 + .01286 + .00002 + 

= (approximately) 9.70824, Ans. 



EXAMPLES. 

3. Given ^64 = 4, ^65 = 4.0207, f 66 = 4.0412, ^67 = 
4.0615 ; find ^66A 

4. Given ^45 = 3.556893, ^47 = 3.608826, ^49 = 3.659306, 
^51 = 3.708430; find ^48. 

5. Given ^5 = 2.23607, yf 6 = 2.44949, ^7 = 2.64575, y/8 
= 2.82843 ; find \'r>M. 

6. (Jivcn the length of a degree of longitude in latitude 
41°=4528 miles; in latitude 42° = 44.59 miles; in latitude 
43°=43.8S miles; in latitude 44° = 43.16 miles. Find the 
length of a degree of longitude in latitude 11' 30'. 



LOGARITHMS. 339 

7. If the amount of $ 1 at 7 per cent compound interest for 
2 years is $ 1.145, for 3 years $ 1.225, for 4 years $ 1.311, and 
for 5 years $ 1.403, what is the amount for 4 years and 6 
months ? 



XLI. — LOGARITHMS. 

444. The logarithm of a quantity to any given base, is the 
exponent of the power to which the base must be raised to equal 
the quantity. 

For example, if a x = m, x is the exponent of the power to 
which the hase, a, must be raised to equal the quantity, m; 
or, x is the logarithm of m to the base a ; which is briefly 
expressed thus : 

x = log a m. 

445. If a remain fixed, and m receive different values, a 
certain value of x Avill correspond to each value of m; and 
these values of x taken together constitute a System of Loga- 
rithms. And as the base, «., may have any value whatever, the 
number of possible systems is unlimited. 

For example, suppose a = 3. 

Then, since 3° = 1, by Art. 444, = log., 1 

" 3 X = 3, " " l = log 3 3 

" 3 2 =9, " " 2 = log 3 9 

Hence, in the system whose base is 3, log 1 = 0, log 3 = 1, 
log 9 = 2, etc. 

Again, suppose a = 12. 

Then, since 12 1 = 12, 1 = log 12 12 

" 12 2 = 144, 2 = log 12 144 

Hence, in the system whose base is 12, log 12 = 1, log 
144 = 2, etc. 



340 ALGEBEA. 

446. The only system in extensive use for numerical com- 
putations is the Common System or Briggs' System, whose 
base is 10. Therefore the definition of the common logarithm 
of a quantity is the exponent of that power of 1.0 which equals 
the quantity. Hence, 

Since 10° = 1, log 10 1 = 

10 1 = 10, log 10 10 = 1 

" 10 2 =100, log 10 100 = 2 

TO 3 = 1000, log 10 1000 = 3 

" 10- 1 = i = .l, log in .l = -l 



a 



10- 2 = ^ = .01, log 10 .01=-2 

'< 10- 3 -^ = .001, log 10 .001 = -3, etc. 

447. It is customary in using common logarithms to omit 
the subscript 10 which denotes the base ; hence, we may write 
the results of Art. 446 as follows : 

log 1 = log .1=- 1 = 9- 10 

log 10 = 1 log .01 = - 2 = 8 - 10 

log 100 = 2 log .001 = -3 = 7 -10 

log 1000 = 3 etc. 

The second form of the results in the second column will be 
found less complicated in the solution of examples. 

448. We infer the following from the first column of 
Art. 447 : 

The logarithm of any number between 1 and 10, lies between 
and 1. 

The logarithm of any number between 10 and 100, lies be- 
tween 1 and 2. 



LOGARITHMS. 341 

The logarithm of any number between 100 and 1000, lies be- 
tween 2 and 3, etc. 

Or, in other words, 

The logarithm of any number with one figure to the left of 
its decimal point, is equal to plus some decimal. 

The logarithm of any number with two figures to the left of 
its decimal point, is equal to 1 plus some decimal. 

The logarithm of any number with three figures to the left 
of its decimal point, is equal to 2 plus some decimal, etc. 

449. Reasoning in the same way from the second column 
of Art. 447, 

The logarithm of any number between 1 and .1, lies between 
and 9 — 10, or between 10 — 10 and 9 — 10. 

The logarithm of any number between .1 and .01, lies be- 
tween 9 — 10 and 8 — 10. 

The logarithm of any number between .01 and .001, lies be- 
tween 8 — 10 and 7 — 10, etc. 

Or, in other words, 

The logarithm of any decimal with no zeros between its 
point and first figure, is equal to 9 plus some decimal — 10. 

The logarithm of any decimal with one zero between its 
point and first figure, is equal to 8 plus some decimal — 10. 

The logarithm of any decimal with two zeros between its 
point and first figure, is equal to 7 plus some decimal — 10, 
etc. 

450. It will be seen from the two preceding articles that 
in general the logarithm of a number consists of two parts, 
one integral, the other decimal. The integral part is called 
the characteristic ; the decimal part, the mantissa. For rea- 
sons which will be given hereafter, only the mantissa of the 
logarithm is given in the tables ; the characteristic must be 
supplied by the reader. The rules for characteristic are based 
on the results obtained in the last parts of Arts. 448 and 449. 



342 ALGEBRA. 

451. I. If the number is greater than 1, the characteristic 
is 1 less than the number of figures to the left of the decimal 

[lit'; lit. 

For example, characteristic of log 354.89 = 2, 

characteristic of log 906328.3 = 5, etc. 

II. If fin' number is less than 1, the characteristic is found 

by subtracting the number of zeros between the decimal point 
cut/ first significant figure from 9; writing — 10 after the 
mantissa. 

For example, characteristic of log .00792 = 7, with — 10 
after the mantissa; characteristic of log .2583 = 9, with —10 
after the mantissa ; etc. 

It is customary in ordinary computation to omit the — 10 
after the mantissa ; it should he remembered, however, that it 
is really a part of the logarithm, and should be allowed for, 
and subjected to precisely the same operations as the rest of 
the logarithm. Beginners will find it useful to write it in 
all cases ; and in some problems it cannot conveniently be 
omitted. 

Note. Many writers, in dealing with the characteristics of the loga- 
rithms of numbers less than 1, combine the two portions of the characteris- 
tic, writing the result as a negative characteristic before the mantissa. 
Thus, instead of such an expression as 7.603582-10, the student will fre- 
quently find 3.6035S2 ; a minus sign being written over the characteristic, 
to denote that it alone is negative, the mantissa being always positive. The 
objection to this notation is the inconvenience of using numbers partly 
positive and partly negative. 

PROPERTIES OF LOGARITHMS. 

452. In any system the logarithm of unity is zero. 
For, since a = 1, for any value of a, = log a 1. 

453. In any system the logarithm of the base itself is 
unity. 

For, since a 1 = a, for any value of a, 1 = log„ a. 



LOGARITHMS. 343 

454. hi any system, whose base is greater than unity, the 
logarithm of zero is minus infinity. 

For, since or 00 = — - = — = 0, — cc = log a 0. 
a go 

If the base is less than unity, the logarithm of is + cc . 

455. In any system the logarithm of the product of any 
number of factors is equal to the sum of the logarithms of 
those factors. 

Assume the equations, 



- "H whence, by Art. 444, [ x ~~ !° g " 



Multiplying, a x X cC> — m n, or a x + \ — m n 

Whence, x + y = log a m n 

Substituting values of x and y, 

log a m n — log a m + log a n. 

If there are three factors, m, n, andj?, 

log a m np = log a (m n Xp) = (Art. 455) log a m n + \og a p 

= log a m + log a n + log a p. 

An extension of this 'method will prove the theorem for any 
number of factors. 

By the application of this theorem, we may find the loga- 
rithm of a number, provided we know the logarithm of each 
of its factors. For example, given log 2 = 0.301030, log 3 = 
0.477121, required log 72. 

log 72 = log (2 x 2 x 2 x 3 X 3) 

= log 2 + log 2 + log 2 + log 3 + log 3 

= 3 x log 2 + 2 x log 3 

= 0.903090 + 0.954242 = 1.857332, Ans. 



344 ALGEBRA. 

EXAMPLES. 

Given log 2 = 0.301030, log 3 = 0.477121, log 7 = 0.845098, 
calculate : 

• 

1. log 48. 4. log 98. 7. log 1G8. 10. log 3087. 

2. log 441. 5. log 84. 8. log 7056. 11. log 15552. 

3. log 56. 6. log 567. 9. log 504. 12. log 14406. 

456. In any system the logarithm of a fraction is equal 
to the logarithm of the numerator minus the logarithm of the 
denominator. 

Assume the equations, 



a x = m) i f x = log a m 

„ > whence, { , oa 

ay — n J (y = \og a n 



Dividing, 


ar m , m 
— = — , or a x ~ v = — 
a^ n n 


Whence, 


x-y = \og a -- 


Substituting 


values of x and y, 




loga — = log a m — log a n, 

IV 



By this theorem, a logarithm being given, we may derive 
certain others from it. For instance, if we -know log 2 = 
0.301030, then 

log 5 = log ^ = i g io - log 2 = 1. - 0.301030 = 0.698970. 

41 



EXAMPLES. 

Given log 2 = 0.301030, log 3 = 0.477121, log 7 = 0.845098, 
calculate : 



LOGARITHMS. 345 



1. log 15. 4. log 175. 7. logTf 

2. log 125. 5. log 3i. 8. log—, 



10 

T 



3. log—. 6. loglH. 9. log5£. 



457. In any system the logarithm of any power of a 
quantity is equal to the logarithm of the quantity, multiplied 
by the exponent of the power. 

Assume the equation, 

a x = in, whence, x = log a m 

Eaising both members of the assumed equation to the^th 

power, 

(a x )P = mP, or aP x = mP 

Whence, px = log„ mP 

Substituting the value of x, 

log a mP=p\og a m. 

458. In any system the logarithm of any root of a quan- 
tity is equal to the logarithm of the quantity, divided by the 
index of the root. 

For, log a v' m = log a (m7) = (Art. 457) - log a m. 

459. In the common system, the mantissa; of the loga- 
rithms of all numbers having the same sequence of figures 
will be the same. 

For example, suppose we know that log 3.053 = .484727. 

Then,log30.53=log(3.053xl0)=log3.053+logl0=.484727 
+ 1 = 1.484727. 

Also, log 30530 = log (3.053 x 10000) = log 3.053 + log 10000 
= .484727 + 4 = 4.484727. 



346 ALGEBRA. 

Again, log.03053=log (^~J=log3.053-logl00=.484727 

-2= .484727 + 8-10 = 8.484727-10. 

It is clear, then, that if a number he multiplied or divided hy 
any integral power of 10, thereby producing another number 
having the same sequence of figures, the mantissse of their 
logarithms will be the same. 

Or, to illustrate, if log 3.053 = .484727, 

then, log 30.53 = 1.484727 log .3053 = 9.48472? - 10 

log 305.3 = 2.484727 log .03053 = 8.484727 - 10 

log 3053. = 3.484727 log .003053 = 7.484727 - 10 

etc. etc. 

We may now see the reason why, as stated in Art. 450, only 
the mantissa? are given in the table ; for if we wish to find the 
logarithm of any number, we have only to find the mantissa 
of the sequence of figures composing it from the table, and can 
prefix the proper characteristic, depending on the position of 
the decimal point, in accordance with the rules stated in Art. 
451. This property of logarithms is only enjoyed by the com- 
mon system, and constitutes its superiority over all others. 

460. Given the logarithm, of a quantity to a certain base, 
to calculate the logarithm of the same quantity to any other 
base. 

Assume the equations, 

a x = m) i (x = log, m 

,,. > whence, < , oa 

O'J =m) ' \y = log 6 m 

From the assumed equations, a x = by 

l I £ 

Hence, (a*)?/ = (tity, or av = b 

Whence, - = lo£„ b 

y 

X 

°h y 



l°g a h 



LOGARITHMS. 347 

Substituting the values of x and y, 

, log„ m 

log,, m = . . 

log„£ 

That is, if we know the logarithm of m to a certain base, a, 

its logarithm to any other base, b, is found by dividing by the 

logarithm of b to the base a. 

461. To show that log a 6 X log 6 a = l, for any values oj 
a and b. 

Assume the equation, 

a x = b, whence x = log a b 

Taking the - power of both members, 

00 

III 
(a x y = b x , or b x = a 

Whence, - = log 6 a 

7 x 

Therefore, log a b X log 6 a = x X - = 1- 

462. We append a few examples to illustrate the applica- 
tions of Arts. 455, 456, 457, and 458. 

e 

L l0g © d= 4 l0g !' (Art. 457) 

= - (log a - log b), (Art. 45G) . 

Co 

2. log % a *V b = log Q a x m \J b) - log % e, (Art. 456) 

= log \f a + log y 7 ^ — log y/ c, (Art. 455) 

= - log a-\ log b log c, (Art. 458). 

n m P 



The following are proposed as exercises 

a b c 
~d~e. 



3. log -7— = log a + log b + log c — log d — log e. 



348 ALGKBBA. 

4. log (J/«xJ 3 X c-) = - log « + 3 log & + ^ log a. 

5. log^ = r, log2--log3. 

3 g o o 

6. log \7 — = - (2 log a — log b — log c). 

V c n 

„ . V a b 1 „ _ _ . 1 

7 - log -^y— = - (log « + log &) - — log c. 

8 - l°g > = t log a — log & — s log c — 2 log tf. 

bc$d 2 4 ° 

( s I a _ ™ \ 1 to 

9 - lo S VV ^ + ( c d ) "J = 5 ( lo S « - log ^) + — (logc+ logd)- 



USE OF THE TABLE. 

463. The table (Appendix) gives the mantissa? of the 
logarithms of all numbers from 1 to 10000, calculated to six 
decimal places. On the first page of the table are the loga- 
rithms of the numbers between 1 and 100. This table is 
added simply for convenience, as the same mantissae are to be 
found in the rest of the table. 

To find the logarithm of any member consisting of four 
figures. 

Find, in the column headed N, the first three figures of the 
given number. Then the mantissa, (if the required logarithm 
will be found in the horizontal line corresponding, in the ver- 
tical column which lias the fourth figure of the given number 
at the top. If only the last four figures of the mantissa are 
found, the first two figures may be obtained from the nearest 
mantissa above, in the same vertical column, which consists of 
six figures. Finally, prefix the proper characteristic (Art. 
451). 



LOGARITHMS. 349 

For example, log 140.8 = 2.148603 

log .05837 = 8.766190 - 10 
log 8516. = 3.930236 

For a number consisting of one or two figures, use the first 
page of the table, which needs no explanation ; for a number 
of three figures, look in the column headed 1ST, and take the 
mantissa corresponding in tbe column headed 0. For exam- 
ple, log 94.6 = 1.975891. 

464. To find the logarithm of a number of more than four 
figures. 

For example, let it be required to find log 3296.78. 

From the table, we find log 3296 = 3.517987 

log 3297 = 3.518119 

That is, an increase of one unit in the number produces an 
increase of .000132 in the logarithm. Then' evidently an in- 
crease of .78 unit in the number will produce an increase of 
.78 X .000132 in the logarithm = .000103 to the nearest sixth 
decimal place; 

Therefore, log 3296.78 = log 3296 + .000103 

= 3.517987 + .000103 = 3.518090, Ans. 

Note. The foregoing method is based upon the assumption that the 
differences of logarithms are proportional to the differences of their corre- 
sponding numbers, which is not strictly correct, but is sufficiently exact 
for practical purposes. 

We derive the following rule from the above operation : 

Find in the table the mantissa of the first four figures, 
without regard to tin' position of the decimal point. 

Find the difference between this and the mantissa of the 
next 7iigher number of four figures ; (called the tabular dif- 
ference, and to be found in the column headed D on each 
page : see Note on page 350.) 

Multiply the tubular difference by the rest of the figures of 
the given number, with a decimal point before them. 

Add the result to the mantissa of the first four figures. 

Prefix the proper characteristic. 



350 ALGEBRA. 

1. Find the logarithm of .02243076. 

Mantissa of 2243 = 350829 

Tabular difference = 194 lj> 

.076 350844 



1.164 
13.58 



Correction = 14.744 = 15 nearly. 

Am. 8.350844-10. 

Note. To find the tabular difference mentally, subtract the last figure 
of the mantissa from the last figure of the next larger, and take the nean si 
whole number ending in that figure to the number in the column headed 
D in the same line. For instance, in finding log .02243076, the last figure 
of the mantissa of 2243 is 9, and of the next larger mantissa, 3 ; 9 from 13 
leaves 4, and the nearest number ending in 4 to 193, the number in the 
column headed D, is 194, the proper tabular difference. 

EXAMPLES. 

Find the logarithms of the following numbers : 





10 < "8 oa "" 1 


.110 \ji. 


"'^ ^""tTlMg XX 


LLJ_U.k/^>Xi 


z> ■ 


2. 


.053. 


6. 


33.0908. 


10. 


912.2:..".. 


3. 


51.8. 


7. 


.0002851. 


11. 


.870092. 


4. 


.2956. 


8. 


65000.63. 


12. 


7303. Oi 8 


5. 


1.0274. 


9. 


.001030741. 


13. 


.0436927 



14. Given log 7.83 = .89376, log 7.84 = .89432; find log 
78309. 

15. Given log .05229 = 8.718419-10, log .05230 = 8.718502 
-10; find log 52.2938. 

16 Given log 315.08 = 2.4984208, log 315.09 =2.4984346; 
find log .003150823. 

17. Given log 18.84 = 1.275081, log 18.87 = 1.275772 ; find 
log .188527. 

18. Given log 9.5338 = .9792660, log 9.5342 = .9792843 ; 
find log 9534071. 



LOGARITHMS. 351 

465. To find the number corresponding to a logarithm. 

For example, let it be required to find the number whose 
logarithm is 3.693845. 

Since the characteristic depends only on the position of the 
decimal point, and in no way affects the sequence of figures 
corresponding, we ought to obtain all of the number corre- 
sponding, except the decimal point, by considering the man- 
tissa only. We find in the table the mantissa 693815, of which 
the corresponding number is 4941, and the mantissa 693903, 
of which the corresponding number is 4 ( .)42. 

Tbat is, an increase of 88 in the mantissa produces an in- 
crease of one unit in the number corresponding. Hence, an 
increase of 30 in the mantissa will produce an increase of §§ of 
a unit in the number, or .34 nearly. Therefore, 

Number corresponding = 4941 + -34 = 4941.34, Ans. 

We base the following rule on the above operation : 

Find in the table the next less mantissa, the four figures 
corresponding, and the tabular difference. 

Subtract the next less mantissa from the given, mantissa. 

Divide the remainder bg the tabular difference / (the quo- 
tient in general cannot be depended upon to more than two 
decimal places.) 

Annex all of the quotient except the decimal point to the 
first four figures of the number. 

Point off. 

Note. The rules for pointing off are the reverse of tire rules for charac- 
teristic given in Art. 451 : 

I. If — 10 is not written after the mantissa, add 1 to the 
characteristic, giving the number of figures to the left of the 
decimal point. 

II. If — 10 is written after the mantissa, subtract the 
characteynstic from 9 ; giving the number of zeros to be 
placed between the decimal point and first figure. 



352 ALGEBRA. 

1. Find the number whose logarithm is 7.950185 — 10. 

950185 
Next less mantissa =950170; four figures corresponding =8916. 

Tabular difference =49) 15.00 (.31 nearly. 

147 

~30 

Therefore, number corresponding = .00891631, Ans. 

EXAMPLES. 

Find the numbers corresponding to the following : 

2. 1.880814. 6. 8.044891-10. 10. 0.990191. 

3. 9.470410-10. 7. 2.270293. 11. 7.115658-10. 

4. 0.820204. 8. 9350064-10. 12. 8.535003-10. 

5. 4.745126. 9. 3.000027. 13. 1.670180. 

14. Given log 113 = 2.05308, log 114 = 2.05690 ; find num- 
ber corresponding to 1.05411. 

15. Given log .08630 = 8.936011 - 10, log .08631 = 8.936061 
— 10 ; find number corresponding to 0.936049. 

16. Given log 2.0702 = .3160123, log 2.0703 = .3160333 ; 
find number corresponding to 9.3160138 — 10. 

17. Given log 548 3 = 2.739018, log 548.9 = 2.739493 ; find 
number corresponding to 7.739416 — 10. 

18. Given log 7.3488 = .8662164, log 7.3492 = .8662401 ; 
find number corresponding to 2.8662350. 

466. In the application of Arts. 455, 456, 457, and 458, we 
have to perform the operations of Addition, Subtraction, Mul- 
tiplication, and Division with logarithms. As some of the 
problems which may arise arc peculiar, wo give a few hints as 
to their solution, which will be found of service. 

1. Addition. If, in the sum, — 10, —20, —30, etc., are 
written after the mantissa, and the characteristic standing be- 



LOGARITHMS. 353 

fore the mantissa is greater than 9, subtract from both parts 
of the logarithm such a multiple of 10 as will make the charac- 
teristic before the mantissa less than 10. 

For example, 13.354802 - 10 should be changed to 3.354802 ; 
28.964316 - 30 should be changed to 8.964316 - 10 ; etc. 

2. Subtraction. In subtracting a larger logarithm from 
a smaller, or in subtracting a negative logarithm from a posi- 
tive, the characteristic of the minuend should be increased by 
10, — 10 being written after the mantissa to compensate. 

For example, to subtract 3.121468 from 2.503964, we write 
the minuend in the form 12.503964 — 10 ; subtracting from 
this 3.121468, we have as a result 9.382496 — 10. 

To subtract 9.635321 — 10 from 9.583427 - 10, we write 
the minuend in the form 19.583427 — 20 ; subtracting from 
this 9.635321 - 10, we have as a result 9.948106 - 10. 

3. Multiplication. The hint already given for reducing 
the result of Addition, applies with equal force to Multiplication. 

To multiply a logarithm by a fraction, multiply first b} r the 
numerator, and divide the result by the denominator. 

4. Division. In dividing a negative logarithm, add to 
both parts of the logarithm such a multiple of 10 as will make 
the quantity after the mantissa exactly divisible by the divisor, 
with — 10 as the quotient. 

For example, to divide 7.402938 — 10 by 6, we add 50 to 
both parts of the logarithm, giving 57.402938 — 60. Dividing 
this by 6, we have as a result 9.567156 — 10. 

EXAMPLES. 

1. Add 9.096004 - 10, 4.581726, and 8.447510 - 10. 

2. Add 7.196070 - 10, 8.822209 - 10, and 2.205683. 

3. Subtract 0.659321 from 0.511490. 

4. Subtract 7.901338 - 10 from 1.009800; 

5. Subtract 9.156243 - 10 from 8.750404 - 10. 



354 ALGEBRA. 



6. Multiply 9.105107 - 10 by 3. 

7. Divide 8.452G33 - 10 by 4. 

8. Divide 9.670392 - 10 by 11. 

9. Multiply 9.6G8311 - 10 by ?. 



SOLUTIONS OF ARITHMETICAL PROBLEMS BY 

LOGARITHMS. 

467. In finding the value of any arithmetical quantity by 
logarithms, we first find the logarithm of the quantity, as in 
Art. 462, by the aid of the table, and then find the number 
corresponding to the result. 

1. Find the value of .0631 X 7.208 X 512.72. 

By Art. 455, log (.0631 x 7.208 x 512.72) = log .0631 
+ log 7.208 + log 512.72 

log .0631= 8.800029-10 
log 7.208= 0.857815 
log 512.72= 2.709880 



Adding, .-. log of Ans. = 12.367724 - 10 

= 2.367724 (Art. 466, 1) 
Number corresponding to 2.367724 = 233.197, Ans. 

„. , „ . . 3368.52 
2. Find the value of -^^g. 

log HJIJH = log 3368.52 - log 7980.04 

log 3368.52 = 13.527439 - 10 (Art. 466, 2) 
log 7980.04= 3.902005 



Subtracting, .-. log of Ans. = 9.625434—10 
Number corresponding =.422118, Ans. 



LOGARITHMS. 355 

3. Find the value of (.0980937) 5 . 

log (.0980937) 5 = 5 x log .0980937 
log .0980937 = 8.991641 - 10 

5 



Multiplying, .-. log of Ans. = 44.958205 - 50 

= 4.958205-10 
Number corresponding = .0000090825, Ans. 

4. Find the value of ^2.36015. 

log ^ 2.3601 5 = * log 2.36015 

log 2.36015 = 0.372940 
Dividing by 7, .-. log of Ans. = 0.053277 

Number corresponding = 1.13052, Ans. 

2 v^5 

5. Find the value of — — . 

3* 

log $ - ] °g 2 + I log 5-5 log 3 
o 

log 2 = 0.301030 

log 5 = 0.698970 ; divide by 3 = 0.232990 

log 3 = 0.477121 0.534020 

Multiply by 5, = 2.385605 ; divide by 6 = 0.397601 
Subtracting, .-. log of Ans. = 0.136419 • 

Number corresponding = 1.36905, Ans. 

Note. The work of the next two examples will be exhibited in the 
customary form, the — 10's being omitted after the mantissse. See Art. 451. 



6. Find the value of ^.00003591. 



356 ALGEBRA. 



log {/ .00003591 = ^ log .00003591 

log .00003591 = 5.555215 
7)5.555215 
log of Ans. Z 9.365031 (Art. 466, 4) 
Ans. = .231756. 



m -n- ", i i r // -032956 \ 

7. Find the value of W ( - 



7.96183/' 



los \J (^SiH) = \ ^ - 03295G " log 7,96183) 



log .032956 = 8.517934 



log 7.96183 = 0.901013 



2)7.616921 
log of Ans. = 8.808460 
Ans. = .0643369. 

Note. In computations by logarithms, negative quantities are used as 
if they were positive ; the sign of the result being determined irrespective 
of the logarithmic work. 

EXAMPLES. 
468. Calculate; by logarithms, the values of the following : 

1. 9.23841 x .00369822. 5. ^3. 

* 

3.70963 x 286.512 g ,g 

1633.72 * ' V 

3. (23.846-t) 8 . 7. ^5. 

4. (- .0009296S7)*. 8. ^.0042937. 



LOGARITHMS. 



357 



18 



9. V- 6829.586. 



112 



10. (1.05624) 



11. (- .0020001G)i£. 



12. 2? x (- 3)*. 



13. 



14. 



3 

5 T 



(-2)* 

3^ 
(- 4) § 



15. m 



ii 



16. V 7239.812. 



17. V .00230508. 



19. 



35 

113 



/ .0872635 U 
\ .132088 / " 



«. i/\- 



22. i> 



23 



21 
13* 



24. f2x('3xf4. 

// 3258.826 \ 
V V 49309.8 ) ' 



25 



/ - 31.6259 W 
' V 429.0162 



27 _ (625.343)- 



(.732465) 



t 



28. 



29. 



30. 



V .000128883 
y. 000827606* 

(_ .746892) ^ 
- (.234521)^ 

ty .00730007 

" * 

(.682913) ^ 



18. V- .000009506694. 31. 



y 5.95463 x V 61.1998 



V 298.5434 
32. (538.217 x .000596899)^. 



33. - 



304.698 x .9026137 



.00776129 X- 16923.24 



34. (18.9503) 11 x (-.213675) 14 . 



358 ALGEBRA. 



A Orro I Q 



35. V 3734.89 x .00001108184. 



36. (2.03172)* x (.712719)* 



y- .00819323 x (.0628513) * 
- .9834171 " 



6/^T7T7777^T- .8/ 



38. \/.035 x V .626671 X V-M721033. 

EXPONENTIAL EQUATIONS. 

469. An Exponential Equation is one in which the un- 
known quantity occurs as an exponent. 

To solve an equation of this form, take the logarithms of 
hoth members according to Art. 457'; the result will be an 
equation which can be solved by ordinary algebraic methods. 

1. Given 31* = 23 ; find the value of x. 
Taking the logarithms of both members, 

log (31*) = log 23 
or, by Art. 457, x log 31 = log 23 

mi lo S 23 1-361728 MWVr - , 

Whence, X = ^ = -^—^ = .91307 < , Ans. 

The value of the fraction '.^~^ may be obtained bv di- 

1.491362 J J 

vision, or better by logarithms, as in Art. 468. 

2. Given .2* = 3 ; find the value of x. 
Taking the logarithms of both members, 

x log .2 = log 3 

log 3 .477121 .477121 

\Y hence, x = 



' log .2 9.301030 — 10 .698970 

We may find the value of the fraction by logarithms exactly 
as if it were positive, and prefix a — sign to the result. Thus. 



LOGARITHMS. 359 

log .477121 = 9.678628 - 10. 
log .698970 = 9.814458 - 10 



Subtracting, = 9.834170 - 10 

Number corresponding = .682606 
Therefore, x = — .682606, Ans. 

EXAMPLES. 

Solve the following equations : 

3. II 1 = 3. 5. 13* = .281. 7. 5*~ 8 = 8 2 * +1 . 

4. .3 r = .S. 6. .703* = 1.09604. 8. 23 3 * + 5 = 31 2 *- 3 . 

APPLICATION OF LOGARITHMS TO PROBLEMS IN 
COMPOUND INTEREST. 

470. Let P = the principal, expressed in dollars. 

Let t = the interval of time during which simple interest 
is calculated, expressed in years and fractions of a }*ear. For 
instance, if the interest is compounded annually, t = 1 ; if 

semi-annually, t = - ; etc. 

Let P = the interest of one dollar for the time t. 

Let n = the number of years. 

Let Ai, A 2 , A 3 , be the amounts at the ends of the 1st, 

2d, 3d, intervals. 

Let A be the amount at the end of n years. 

Then A l = P + PP = P(l + P) 
A 2 = A 1 + A 1 B = A 1 (1 + R) 

=p (i + p) (i + p) = p (i + py 

A 3 = A, + A 2 P = A 2 (1 + P) 

= p(i + py 2 (i + p) = p(i + py 



300 ALGEBRA. 

71 

As there are - intervals, the amount at the end of the last, 
v 

according to the law observed above, 

A =* P (1 + Sp. 

1. Given P, t, R, and n, to find A. 

n 

As A = P (1 + R) *., we have by logarithms, 
log A = log P (1 + Rp = log P + log (1 + Rp 

71 

= log P + - log (1 + R). 

h 

Example. What will be the amount of $7,325.67 for 3 
years 9 months at 7 per cent compound interest, the interest 
being compounded quarterly ? 

Here P = 7325.67, t = j , R — .0175, n = 3f, - == 15. 



log P = 3.864848 

i 



log (1 + R) = 0.007534 ; multiply by 15 = 0.113010 
Adding, . •. log of A = 3.977858 

Number corresponding, A = $ 9502.93, Ans. 

2. Given t, R, n, and A, to find P. 

n A 

As A = P (1 + R)t , .-. P = ; or, by logarithms, 

(1 + R)~ 

/log P = log A — log (1 + Rp = log A — " log (1 + R). 

Example. What sum of money will amount to $ 1 ,76« \.5B at 
5 per cent compound interest in 3 years, the interest being 

compounded semi-annually ? 



LOGARITHMS. 361 

1 n 

Here t = % , B = .025, » = 3, A = 1763.55, - = 6. 

log ^ = 3.246388 
log (1 + P) = 0.010724; multiply by 6 = 0.064344 

Subtracting, .-. log P = 3.182044 

Number corresponding = f> 1520.70, Ans. 

3. Given P, t, P, and A, to find n. 
In Art. 470, 1, we sbowed that 

log^ = logP + -log(l + P) 

1/ 

.•.^log(l + P)=log^-logP 

£ (log A — log P) 
* • l ~ log (1 + 5) • 

Example. In bow many years will $300.00 amount to 
8 400.00 at 6 per cent compound interest, the interest being 
compounded quarterly ? 

Here P = 300, t = | , P = .015, A = 400. 

log 400 - log 300 2.602060 - 2.477121 .124039 

.'.71 = 



4 log 1.015 ~ 4 x .006466 .025864 

= 4.83 years, Ans. 

4. Given P, t, n, and A, to find P. 

n 
We sbowed, in Art. 470, 3, that - log(l + P) = log^-logP 

V 

1 /1 , T>\ log^ — logP 

••• log (1 + P) ■ • 

Example. If $ 500.00 at compound interest amounts to 
$689.26 in 6 years and 6 months, the interest being com- 
pounded semi-annually, what is tbe rate per cent per annum '? 



G2 



ALGEBRA. 



1 n 

Here P = 500, t = 7> ,n = 6h,A = 689.26, - = 13. 

z t 

, ,- ™ log 689.26- log 500 



13 



log 689.26 = 2.838383 
log 500 = 2.698970 



Subtracting, 



= 0.139413 
Dividing by 13, .-. log (1 + B) = 0.010724 

Number corresponding = 1.025 = 1 + II, or R = .025. 

That is, one dollar gains $ .025 semi-annually ; or the rate is 
5 per cent per annum. 



EXPONENTIAL AND LOGARITHMIC SERIES. 
471. We know that for any values of n and x, 



1 + 



x / 1 \ n 

= ( 1 + n, 



Expanding by the Binomial Theorem, we obtain 



1 n (n - 1) 1 n (n - 1) (n - 2) 1 

1 + n - H r?i 5 H rr; ? + 



w 



w." 



[3 



?r 



. 1 v x (nr—1) 1 ?? x(nx— V)(nx— 2) 1 

w 2 ?r 3 ?r 



+ 



or, 



!_! (l-l)(!- 2 ) 

T77- T^ + 



L3 



x \x x \x 

V n I n 

= 1 + x H r^ + 



11 



\1 



; + 



LOGARITHMS. 363 

This is true for all values of n ; hence, it is true however 
large n may he. Suppose n to he indefinitely increased. Then 

1 2 

the limiting values of the fractions - , — , etc., are (Art. 210). 

n n 

Hence, at the limit, we have, 
1 1 



1 + 1 + 777 + T7V + 



-f^ + Sr + 



[3_ J ' ' [2 ' [3 

The series in the bracket we denote by e ; hence, 



2 3 

x x 



<? = ! + * + - + - + 



472. To expand a x in poivers of x. 

Let a = e m ; whence (Art. 444), 7ft = log c ft. 



m 2 a? 2 ra 3 x 3 



Then a* = e ra x = (Art. 471) 1 + m a; + -y^- + -r^- + 



Substituting the value of m, 



V 2 /I \ *\ 



a* = 1 + (l0g e ft) X + (log, ft) 2 — + (log e ft) 3 j^ + 

This result is called the Exponential Theorem. 

473. The system of logarithms which has e for its base, 
is called the Napierian System, from Napier, the inventor of 
logarithms. The value of e may be easily calculated from the 
series of Art. 471, and will be found to be 2.7182818 

474. To expand log,, (1 + x) in poivers ofx. 

a* = {l+(a-l)}* = l + x(a l -l).+ X(x ~ 1) (a-iy 

+ g (s-l)(s-2) + 



= l + x {(ft- 1) - - ( -^ + ^^ } + terms con- 

taining x 2 , x s } etc. 



364 ALGEBRA. 

But (Art. 472), a x = 1 + x (log e a) + terms containing x% 

As the two values of a x are equal for all values of x, by the 
Theorem of Undetermined Coefficients the coefficients of x in 
the two expressions are equal; hence, 

log e a=(a — l) ^ 1 3 

Putting a = 1 + x, and therefore a — 1 = x, we obtain 

Syt" /ytO 

log e (l + «)=» — — + — — 



Note. This formula might be used to calculate Napierian logarithms ; 
but unless x is a very small fraction, the series in the second number is 
either divergent or converges very slowly, and hence is useless in most 
cases. 

475. To obtain a more convenient formula for calculating 
the Napierian logarithm of a number. 

X 2 X s X* X 5 



By Art. 4,4, log e (1 + x) = x — y + y — y + y — 

put X = — X, 



X 2 X 3 X* X s 



.-.log e (l — x)=—x 



Subtracting, 



2 x 3 2x h 
• . log e (1 + x) - log, (l-x) = 2x+ — +- ir + 



or, by Art. 456, log,, f j 



1 + x\ c ( x 3 x 5 

> = 2 [x+- 7r +- T r + 



Let x = 



x) 3 5 

1 



1 + 



2n + l 
1 



l+x_ 2n + l _ 2n + 1 + 1 _ 2n + 2 __ n + l 

l—x~~7 ~T~ ~2n + l — l~ 2 u n 

2n + l 



LOGARITHMS. 365 

Substituting, .\ log, p— J = lo ge (n + 1) — log e » 

~~ \2n + 1^3(2n + iy ^ 5 (2 n + l) 3 T " / 

■■■^g e (n+l)=\o ge n+2(^ + 3 ^ +1) ,+ 5{:Jn+1)b + ) 

476. To calculate log e 2, put n = 1 in the formula of Art. 
475. 

/ 1 1 1 \ 

...log e 2 = log e H-2^2 TI +3 (2 + 1) s+5^2q:-i)5+ j 

or, since log e 1 = 0, 

/l 1 1 1 1 1 

log. 2 = 2 ^3 + 81 + 12i5 + 15309 + 177147 + 1948617" 1 """ 



= 2 (.3333333 + .0123457 + .0008230 + .0000653 
+ .0000056 + .0000005 + ) 

= 2 x .3405734 = .6931468 = .693147, correct to the 
sixth decimal place. 

From log,, 2, we may calculate log e 3 ; and so on. We shall 
find log e 10 = 2.302585. 

477. Tn calculate the common logarithm of a number from 
its Napierian logarithm. 

By Art. 460, changing b to 10, and a to e, we obtain 

"»■ " = feS = &3S3«i b ~ » = - 434 - 0945 X ,0g ' * 
For instance, log I0 2 = .4342945 x .693147 = .301030. 

The multiplier by which logarithms of any system are de- 
rived from the Napierian system, is called the modulus of that 
system. Hence, .4342945 is the modulus of the common sys- 
tem. 



366 ALGEBRA. 

As tables of common logarithms are met with more fre- 
quently than tables of Napierian, a rule for changing common 
logarithms into Napierian may be found convenient. 

RULE. 

Divide the common logarithm by .4342945. 

For example, to find the Napierian logarithm of 586.324, 

common log 586.324 = 2.768138 

Divide by .4342945, .-. Napierian log 586.324 = 6.373873, Ans. 

Another method would be to multiply the common logarithm 
by 2.302585, the reciprocal of .4342945. 

Napierian logarithms are sometimes called hyperbolic loga- 
rithms, from having been originally derived from the hyper- 
bola. They are also sometimes called natural logarithms, 
from being those which occur first in the investigation of a 
method of calculating logarithms. Napierian logarithms are 
seldom used in computation, but occur frequently in theoretical 
investigations. 



ARITHMETICAL COMPLEMENT. 

478. The Arithmetical Complement of the logarithm of 
any quantity is the logarithm of the reciprocal of that quantity. 

For example, if log 4098 = 3.612572, then 

ar. co. l#g 4098 = log — — = log 1 - log 4098 
° 4098 ° ° 

= - 3.612572 = 6.387428 - 10. 
Again, if log .06689 = 8.825361 - 10, then 

ar. co. log .06689 = log --?— - = - (8.825361 - 10) 
= 10 - 8.825361 = 1.174639. 



LOGARITHMS. 367 

The following rules will be evident from the preceding 
illustrations : 

To find the arithmetical complement of a positive loga- 
rithm, suit nut it from 10, writing — 10 after the mantissa. 

To find the arithmetical complement of a negative loga- 
rithm, subtract that portion of it besides the — 10 from 10. 

The only application of this is to exhibit the work of calcu- 
lation by logarithms in a more compact form in certain cases. 
It depends on the' principle that subtracting a logarithm or 
adding its arithmetical complement gives the same result. 

For, suppose we are to calculate by logarithms. 

. a X b . ( . 1 1 

l0 «7^ = l0g l ax( ' x c x ( z 



= log a + log b + log - + log 



1 i 1 

c + ] ° S d 



= log a -f- log b + ar. co. log c + ar. co. log d. 

That is, the work can be exhibited in the form of the addi- 
tion of four logarithms, instead of the subtraction of the sum 
of two logarithms from the sum of two others. The principle 
is only applicable to the case of fractions ; and the rule to be 
used is, 

Add together the logarithms of the quantities in the numer- 
ator, and the arithmetical complements of the logarithms of 
the- quantities in the denominator. 

Example. Calculate the value of „ 't, — _1 00 . 
1 613.8 x .0* .23 

log Qi^xljfn = l0 S 79 ' 23 + lQ g 10 - 39 + ar " co - Io S 613 - 8 

+ ar. co. log .07723 



368 ALGEBRA. 

log 79.23= 1.898890 

log 10.39 = 1.016616 

ar. co. log G13.8 = 7.211973 - 10 

ar. co. log .07723= 1.112211 

Adding, .-.log of Ans. = 11.239693 - 10 = 1.239693 
Number corresponding = 17.3657, Ans. 

Note. The arithmetical complement may be calculated mentally from 
the logarithm, by subtracting the last significant figure from 10, and all the 
others from 9. 



MISCELLANEOUS EXAMPLES. 
479. 1. Find log 3 2187. (See Art. 111.) 

2. Find log 5 15625. 

3. Find the logarithm of -rr to the base —2. 

* 61 

4. Find the logarithm of — to the base 8. 

5. Find the characteristic of log 2 183. 

6. Find the characteristic of log 5 4203. 

7. Given log 2 = .301030, how many digits are there in 
2 17 ? 

8. Given log 3 = .477121, how many digits are there in 
3^? 

9. Findlog 13 56. (See Art. 460.) 

10. Find log 8 163. 

11. Find loggo 411. 

12. What sum of money will amount to § 8705.50, in 7 
years, at 7 per cent compound interest, the interest being com- 
pounded annually ? 



GENERAL THEORY OF EQUATIONS. 360 

13. In how many yours will a sum of money double itself at 
6 per cent compound interest, the interest being compounded 
semi-annually ? 

14. What will be the amount of $1000.00 for 38 years 
3 months, at 6 per cent compound interest, the interest being 
compounded quarterly ? 

15. At what rate per cent per annum will $2500.00 amount 
to $ 3187.29 in 3 years and 6 months, the interest being com- 
pounded quarterly ? 

16. In bow many years will 8 9681.32 amount to $ 15308.70 
at 5 per cent compound interest, the interest being compounded 
annually ? 

17. Using the table of common logarithms, find the Na- 
pierian logarithm of 52.9381 (Art. 477). 

18. Find the Napierian logarithm of 1325.07. 
19 Find the Napierian logarithm of .085623. 
20. Find the Napierian logarithm of .342977. 



XLII. — GENERAL THEORY OF EQUATIONS. 

480. The general form of a complete equation of the nth 
degree is 

x n -\- 1~> x n ~ l + q x n ~' 2 + + t x 2 + u x + v = 

"Where n is a positive integer, and the number of terms is n + 1. 
The quantities j), q, t, u, v are either positive or nega- 
tive, integral or fractional ; and the coefficient of x 11 is unity. 

481. In reducing an equation to the general form, all the 
terms must be transposed to the first member, and arranged 
according to the powers of x. If as" has a coefficient, it may 
be removed by dividing the equation by that coefficient. 



370 ALGEBRA. 

482. A Root of an equation is any real or imaginary ex- 
pression, which, being substituted for its unknown quantity, 
satisfies the equation, or makes the first member equal to 
(Art. 166). 

We assume that every equation has at least one root. 

483. An equation of the third degree containing only one 
unknown quantity, or one in which the cube is the highest 
power of the unknown quantity, is usually called a cubic equa- 
tion. 

484. An equation of the fourth degree containing only one 
unknown quantity is usually called a biquadratic equation. 



DIVISIBILITY OF EQUATIONS. 

485. If a is a root of an equation in the form 

x n +qi x n ~ l + q x n ~ 2 + + tx 2 + ux + V — 0, 

then the first member is divisible by x — a. 

It is evident that the division of the first member by x — a 
maybe carried on until x disappears from the remainder. Let 
Q represent the quotient, and R the remainder, which is inde- 
pendent of x ; then the given equation may be made to take 

the form 

. (x - a) Q + B = 0. 

But if x = a, then (x — a) Q = 0, and, consequently, 

R = 0; 

that is, x — a is a factor of the first member of the given equa- 
tion, as it is contained in it without a remainder. 

486. Conversely, if the first member of am. equation in the 
form 

* 

x n -\-px n ~ 1 -\- qx n ~ 2 + + t x 2 + ux + v = 

is divisible by x — a, then a is a root of the equation. 



GENERAL THEORY OF EQUATIONS. 371 

For, if the first member of the given equation is divisible by 
x — a, then tbe equation may be made to take the form 

(x — a) Q = ; 

and it follows from Art. 330 that a is a root of this equation. 

EXAMPLES. 

By the method of Art. 486, 

1. Prove that 3 is a root of the equation 

x 3 — 6 x 2 + 11 x — 6 = 0. 

2. Prove that — 1 is a root of the equation x 3 + 1 = 0. 

3. Prove that 1 is a root of the equation 

x 3 + x 2 — 17 a; + 15 = 0. 

4. Prove that — 2 is a root of the equation 

a; 4 — 3 x 2 + 4 x + 4 = 0. 

5. Prove that 4 is not a root of the equation 

x* — 5 x 3 + 5 x 2 + 5 x - 6 = 0. 

NUMBER OF ROOTS. 

487. Every equation of the nth degree, containing but one 
unknown quantity, has n roots, and no more. 

Let a be a root of the equation 

x n +2) x n ~ x + qx n ~ 2 + + tx 2 + ux + v — 0; 

then, by Art. 485, the first member is divisible by x — a, and 
the equation may be made to take the form 

(x — a) (V 1 - 1 + 2h x n ~ 2 + + «j x + i\) = 0. 

The equation may be satisfied by making either factor of 
the first member equal to (Art. 330) ; hence, 

x • — a = 

and x»~ 1 +p 1 x n - i + +u l x + v 1 = 0. (1) 



372 ALGEBRA. 

But equation (1) must have some root, as b, and may be 
placed under the form 

(x — b) (x 11 - 2 +jh x n ~ 3 + + v,x + v,)=0', 

which is satisfied by placing either factor of the first member 
equal to ; and so on. 

Since each of the factors x — a, x — b, etc., contains only 
the first power of x, it is evident that the original equation can 
be separated into as many such binomial factors as there are 
units in the exponent of the highest power of the unknown 
quantity, and no more ; that is, into n factors, or 

(x — a) (x — b) (x — c) (x — T) = 0. 

Hence, by Art. 330, the equation has the n roots a, b, c, /. 

Moreover, if the equation had another root, as r, then it 
must contain another factor x — r, which is impossible. 

488. It should be observed that the n binomial factors of 
which the general equation of the nth degree is composed, are 
not necessarily unequal j hence, two or more of the roots of an 
equation may be equal. Thus, the equation 

x * _ g x 2 + 12 x _ 8 = 

may be factored so as to take the form 

(x-2)(x-2) (.r-2)=0, or (*-2) 3 = 0; 
and hence the three roots are 2, 2, and 2. 

489. It will be readily seen that any equation, one of 

whose mots is known, may be depressed to another of the next 
lower degree, which shall contain the remaining roots. Hence, 
if all the roots of an equation are known excepl two, those 
may be obtained from the depressed equation, hy the rules for 
quadratics. 

1. One root of the equation x s + 2 x 1 — 23 x — 60 = is — 3 ; 
what are the others ? 



GENERAL THEORY OF EQUATIONS. 373 

Dividing x 3 + 2 x~ — 23 x — 60 by x + 3, the given equation 
may be put in the form 

O + 3)O 2 -x-20)=0. 

Thus,- the depressed equation is x 2 — x — 20 = 0. 

Solving this by the rules for quadratics, we obtain x = 5 or 

— 4 ; which are the remaining roots. 

EXAMPLES. 

2. One root of the equation x z — 19 x + 30 = is 2 ; what 
are the others ? 

3. Eequired the three roots of the equation X s = a 3 , or 

x 3 — a 3 — 0. 

4. One root of the equation x 3 + x 2 — 16 x + 20 = is — 5 ; 
reqxiired the remaining roots. 

5. Two roots of the equation x A — 3 x 3 — 14 x" + 48 x — 32 = 
are 1 and 2 ; required the remaining roots. 

6. One root of the equation x 4 — 7 x 3 + 3 x + 3 = is 1 ; 
what equation contains the remaining roots ? 

7. One root of the equation 6 x 3 — x~ — 32 x + 20 = is 2 ; 
what are the others ? 

8. Two roots of the equation 20 x* - 169 x 3 + 192 x~ + 97 x 

— 140 = are 1 and 7 : what are the others ? 



5 



FORMATION OF EQUATIONS. 

490. An equation having any given roots may be formed 
by subtracting each root from the unknown quantity, and pla- 
cing the product of these binomial factors equal to 0. 

For it is evident, from principles already established, that 

an equation having the n roots a, b, c, I may be written 

in the form 

(x — a) (x — b) (x — c) (x — I) = 0. 



374 ALGEBRA. 

After performing the multiplication indicated, the equation 
will assume the form . 

x n +2?x n ~ 1 + qx n ~ 2 + + tx 2 + ux + v = 0. 

(Compare Art. 329.) 

1. Form the equation whose roots are 1, 2, and — 4. 
Result, (x — 1) (cc — 2) (cc + 4) = 

or, x 3 + x 2 — 10 x + 8 = 0. 

EXAMPLES. 

Form the equations whose roots are : 

2. - 1, - 3, and - 5. 6. 1, 2, 3, and 4. 

3. 5, - 2, and - 3. 7. 4, 4, and 5. 

4. 1, - , and - . 8. 0, - 1, 3, and 4. 

5. ± 1 and ±2. 9. - 5 ? , - 2, and ? 

4 3 



COMPOSITION OF COEFFICIENTS. 

491. The coefficient of the second term of an equation of the 
nth degree in its general form is the sum of all the roots with 
their signs changed; that of the third term is the sum of their 
products, taken two and two ; that of the fourth term is the 
sum of their 'products, taken three and three with their .signs 
changed, etc. ; and the last term is the product of all the roots 
with their signs changed. 

For, resuming the equation 

(x '— a) (x — b)(x — c) (x — k) (x — l)=0, 

if we perform the multiplication indicated, we obtain 

(x — a) (x — b) = x~ — (a + b) x + a b, 
(x—a)(x — b) (x—c) = x 3 —(a + b+c)x+(ab + ac+bc)x—abc, 



GENERAL THEORY OF EQUATIONS. 375 

and so on. When n factors have heen multiplied, the coeffi- 
cients of the general equation become 

jp = — a — b —c— — k — I 

q = ab + ac-\-bc-\- + kl 

r = — a b c — ab d — a c d — — ikl 



v = ±a b c kl 

which corresponds with the enunciation of the proposition ; 
the upper sign of the value of v being taken when n is even, 
and the lower sign when n is odd. 

492. If ^=0, that is, if the second term of an equation 
be wanting, the sum of the roots will be 0. 

If v = 0, that is, if the absolute term of an equation be want- 
ing, at least one root must be 0. 

493. Every rational root of an equation is a divisor of the 
last term. 

494. When all the roots of an equation but two are known, 
the coefficient of the second term of the depressed equation 
(Art. 489) can be found by subtracting the sum of the known 
roots, with their signs changed, from the coefficient of the 
second term of the original equation. The absolute term of 
the depressed equation can be found by dividing the absolute 
term of the original equation by the product of the known 
roots with their signs changed. 

EXAMPLES. 

Find the sum and product of the roots in the following : 

1. a; 3 -7a;+6 = 0. 2. 2 x*- 5 x s - 17 x 2 + 14 x + 24 = 0. 

In the following example obtain the depressed equation by 
the method of Art. 494 : 

3. Two roots of the equation x 4 — 5 x s — 2 x 2 + 12 x + 8 = 
are 2 and — 1 ; what are the others ? 



376 ALGEBRA. 



FRACTIONAL ROOTS. 

495. An equation whose coefficients are all integral, the 
coefficient of the first term being unity, cannot have a 
rational fraction as a root. 

If possible, let -, a rational fraction in its lowest terms, be 
a root of the equation 

x n + px n ~ l + qx n ~ 2 + + tx 2 + UX + v = 0, 

where p, q, , t, u, v are integral. Then 

a\ n faV' 1 fa\ n - 2 ( a\ 2 fa 



Multiplying through by lf~ l , and transposing, 

- n = — ( 2 ,a n - 1 +qa n - 2 b + + ta 2 b n ~ s + ua u n ~ 2 + vb n ~ l ). 

Now, as - is in its lowest terms, a and b can have no com- 
mon divisor ; therefore a n and b can have no common divisor ; 

a" 
hence — is in its lowest terms. Thus, we have a fraction in 
o 

its lowest terms equal to an entire quantity, which is impossi- 
ble. Therefore no root of the equation can be a rational 
fraction. 

Note. The equation may have an irrational fraction as a root, such as 

' for example. Such a root, whose value can only be expressed 

4 
approximately by a decimal fraction, is called incommensurable. 



IMAGINARY ROOTS. 

496. If the coefficients of an equation be real quantities, 
imaginary roots enter it by pairs, if at all. 



Suppose a + b ^ — 1 to be a root of the equation 

X n + p X" ~ l + q X n ~ 2 + + t X 2 + U X + V = 0. 



GENERAL THEORY OF EQUATIONS. 377 



Substituting a + b y/— 1 for x, and developing each expres- 
sion by the Binomial Theorem, all the odd terms of each series 
will contain either powers of a, or even powers of b y/ — 1, and 
are therefore real ; while all the even terms contain the odd 
powers of b y — 1, and are therefore imaginary. Representing 
the sum of all the real quantities by P, and the sum of all 
the imaginary quantities by Q y — 1, we have 



P+ £y/-l=0. 

This equation can be true only when both P and Q equal 0. 

If we now substitute a — b y/ — 1 for x, we find that the 
series differ from the former only in having their even or 
imaginary terms negative. Hence, we obtain as the first 
member 



P-Q^-l, 

which must be equal to 0, for we have already shown that both 
P and Q equal 0. Thus, a — b y/— 1 satisfies the equation. 

Similarly, we may show that, if b y/ — 1 is a root of the equa- 
tion, then will — b y/— 1 also be a root of the equation. 

497. The product of a pair of imaginary quantities is 
always positive. Thus, 



(a + 6y/-l) (a-b^-l) = a 2 + b 
and y/^l) (- b y/^I) = b\ 



TRANSFORMATION OF EQUATIONS. 

498. To transform an equation into another which shall 
have the same roots with contrary signs. 

Let the given equation be 

x n + p x n ~ 1 + q x n ~ 2 + + t x 2 + u x + v = 0. 



378 ALGEBRA. 

Put x = — y; then whatever value x may have, y will have 
the same value with its sign changed. The equation now 
becomes 

{-y) n +p(-y) n - l + q(-y) n - 2 + + t(-y) 2 + u{- y ) 

+ v = o. 

If 11 is even, the first term is positive, second term nega- 
tive, and so on ; and the equation may be written 

yn_ p yU-\ + (1 yn~1_ + f y'2 _ ^ y _J_ y _ Q Qj 

If n is odd, the first term is negative, second term positive, 
and so on ; hence, changing all signs, we write the equation 

y" —v y n ~ 1 + 1 y n ~ 2 — — ^ y 1 + u y~ v — o. (2) 

From (1) and (2) it is evident that to effect the desired 
transformation we have simply to change the signs of the 
alternate terms, beginning with the second. 

Note. The preceding rule assumes that the given equation is complete 
(Art. 300) ; if it be incomplete, any missing term must be put in with 
zero as a coefficient. 

1. Transform the equation x 3 — 7x + 6 = into another 
which shall have the same roots with contrary signs. 

We may write the equation x 3 + . ar — 7 x + 6 = 0. 

Applying the rule, 

x s _ o . x 2 — 7 x — 6 = 0, or x z — 7 x — 6 = 0, Ans. 

EXAMPLES. 

Transform the following equations into others which shall 
have the same roots with contrary signs: 

2. x * - 2 x s + x - 1.32 = 0. 3. x 5 -3x 2 + 8 = 0. 

499. To transform an equation into another whose roots 
shall be some multiple of those of the first. 

Let the given equation he 

x n +px n - 1 + qx n ~ 2 + + tx' 2 + ux + v = 0. 



GENERAL THEORY OF EQUATIONS. 379 

v 
Put x = — ; then whatever value x may have, y will have 
m 

a value m times as great. The equation now becomes 

(i)r + jjLY-\Ji\"-\ +,(£)*+.(»)+.=* 

Multiplying through by ra n , we have 
y n +pmy n ~ 1 +qm 2 y n ~ 2 + + tm n ~ 2 y 2 + um n ~ 1 y + vm" = 0. 

Hence, to effect the desired transformation, multiply the 
second term by the given factor, the third term by its square, 
and so on. 

Similarl}'-, we may transform an equation into one whose 
roots shall be those of the first divided by some quantity. 

1. Transform the equation x 3 — 7 x — 6 = into another 
whose roots shall be 4 times as great. 

The equation may be written, x s -f- . x 2 — 7 x — 6 = 0. 

Then, by the rule, 

x 3 - 4 2 . 7 x - 4 3 . 6 = 0, or x 3 - 112 x - 384 = 0, Ans. 

EXAMPLES. 

2. Transform the equation x z — 2 x 2 + 5 = into another 
whose roots shall be 5 times as great. 

3 x s 

3. Transform the equation x i -| — 27 = into another 

whose roots shall be one third as great. 

500. Ta transform an equation containing fractional 
coefficients into another whose coefficients are integral, that of 
the first term being unity. 

If in Art. 499 we assume m equal to the least common 
multiple of the denominators, it will always remove them ; 
but often a smaller number can be found which will produce 
the same result. 



380 ALGEBRA. 

1. Transform the equation x s — — — — + — — = into 

o So lUo 

another whose coefficients shall he integral. 

The least common multiple of the denominators is 108 ; so 
that one solution would he, by Art. 499, 

rf_M8. |-.M8-:| + 108^ = 0. 

An easier way, however, is as follows ; the denominators 
may be written 3, 3 2 X 2' 2 , and 3 3 X 2 2 , so that the multiplier 
3 x 2 or 6 will remove them. Hence, by Art. 499, we have 

x a_ 6 * a __ 6 3 * +6 s * =0 or x *-2x*-x + 2 = 0, 

ob 106 

whose roots are 6 times as great as those of the given 
equation. 

EXAMPLES. 

Transform the following equations into others whose coef- 
ficients shall be integral : 

**'+T~ 7 4= - *.- , + re-g-»= a 

av-^ + fwb. 5.^-5^-^ + 5 = 0. 

no 4 I 

501. To transform an equation into another ivhose roots 
shall be the reciprocals of those of the first. 

Let the given equation be 

x n + p x n ~ 1 + q x n ~ - + + t x 2 + u x + v = 0. 

Put x = - ; then whatever value x may have, y will be its 

v 

reciprocal. The equation now becomes 

1 p q t u n 

y y y y y 



GENERAL THEORY OF EQUATIONS. 381 

Multiplying through by y", and reversing the order, 

vy n + uy n ~ 1 + ty"- 2 + + qy 2 +py + 1 = 0. 

Dividing through by v, 

u , t a „ 7} 1 

n + _ y «-i + - gf-» + + Lf + L l/+ ± =0 . 

v « w «; «; 

Hence, to effect the transformation, write the coefficients in 
reverse order, and then divide by the coefficient of the first 
term. 

EXAMPLES. 

Transform the following equations into others whose roots 
shall he the reciprocals of those of the first : 

1. a; 3_6x 2 +ll iC -6 = 0. 3. x 3 -9x 2 + — -i = 0. 

7 49 

2. x i ~x 3 -3x 2 + x + 2 = 0. 4. x 3 -4x 2 + 9 = 0. 

502. To transform an equation into another whose roots 
shall differ from those of the first by a given quantity. 

Let the given equation be 

s"+ju""' + qx n ~ 2 + + tx* + ux + v = 0. (1) 

Put x = y + r, and we have 

(ff + r)*+p(t,. + r)— 1 + + u (y + r ) + v=0. (2) 

Developing (y + r) n , (y + r)"-\ , by the Binomial The- 
orem, and collecting terms containing like powers of y, we 
have an equation of the form 

yn +pi yn-l + ^ y «-1 + + ^ y 2 + ^ y + ^ = Q ( o) 

As y=x — r, the roots of (3) are evidently less by r than 
those of (1). By putting x = y — r, we shall obtain in the 
same way an equation whose roots are greater by r than 
those of (1). 

503. If n is small, the operation indicated in Art. 502 
may be effected with little trouble; but for equations of a 
higher degree a less tedious method is better. 



382 ALGEBRA. 

If in (3) we put y = x — r, we shall have 

(x—rf +2h(x — r) n - 1 + +u 1 (x — r) + v 1 = 0, (4) 

which is, of course, identical with (1), and must reduce to (1) 
when developed. If we divide (4) by x — r, we obtain 

(x - r)" ~ J +jh - r) n - 2 + q i {x-r)"-"+ +u, (5) 

as a quotient, with a remainder of v v Dividing (5) by x — r, 
we obtain a remainder of u x ; and so on, until we obtain all 
the coefficients of (3) as remainders. 

Hence, to effect the desired transformation, 

Divide the given equation by x — r or x + r> according as 
the roots of the transformed equation are to be less or greater 
than, those of the first by r, and the remainder will be the 
absolute term of the transformed equation. Divide the quo- 
tient just found by the same divisor, and the remainder will 
be the coefficient of the last term but one of the transformed 
equation ; and so on. 

504. 1. Transform the equation x 3 + 3 x 2 — 4^ + 1 = 

into one whose roots shall be greater by 1. 

Using the method of Art. 502, put x = y — 1. 
Then, (y-iy + 3(y~iy-4(y-l) + l = 0, 
or, t/ 3 - 3 y 2 + 3 7/ - 1 + 3 / - 6 y + 3 - 4 y + 4 + 1 = 0, 
or, if — 7 y + 7 = 0, Ans. 

EXAMPLES. 

2. Transform the equation x 3 — x — 6 = into one whose 
roots shall be less by 8. 

3. Transform the equation x* -f 6 x 3 — x 1 — 5 x — 1 = 
into one whose roots shall be greater by 3. 

505. To transform a complete equation into one tchos* 
second term shall be wanting. 



GENERAL THEORY OF EQUATIONS. 383 

The coefficient of y" -1 in (2), Art. 502, is n r + p. Hence, 
in (3), p y = n r + p. To make ^ = 0, it is only necessary to 

make nr + p =0, or r = — — ; hence, to effect the desired 

n 

V 
transformation, put x = y — - ; that is, put x equal to y } 

minus the coefficient of the second term of the given equation 
divided by the degree of the equation. 

1. Transform the equation x 3 — G x~ -j- 9 x — G = into 
another whose second term shall be wanting. 

Here p = — 6, n = 3 ; then, put x = y — — — = y + 2. 

Result, (y + 2) 3 - 6 (y + 2) 2 + 9 (y + 2) - 6 = 0, 

or, tf + 6y* + 12y + 8-6y*-24:y-24:+9y +18-6 = 0, 

or, y 3 — 3 y — 4 = 0, Ans. 

EXAMPLES. 

Transform the following equations into others whose second 
terms shall be wanting : 

2. x 2 -px + q = 0. 4. x 3 + 6x 2 -3x + 4: = 0. 

3. x s + x 2 + 4 = 0. 5. x 4 - 4 x 3 - 5 x — 1 = 0. 



DESCARTES' RULE OF SIGNS. 

506. A Permanence of sign occurs when two successive 
terms of a series have the same sign. 

A Variation of sign occurs when two successive terms of a 
series have contrary signs. 

DESCARTES' RULE. 

507. A complete equation cannot have a greater number 
of positive roots than it has variations of sign, nor a greater 
number of negative roots than it has permanences of sign. 



384 ALGEBRA. 

Let any complete equation have the following signs : 

+ + + - + - + 

in which there are three permanences and five variations. 

If we introduce a new positive root a, we multiply this by 
x — a (Art. 490). Writing only the signs which occur in the 
operation, we have 



+ + + - + 


— 1 


+ — 




+ + + - + 


- + — 


+ 


- + — + + 



+ ± ± — + — + — ± + 

123456789 10 

a double sign being placed wherever the sign of a term is 
ambiguous. 

However the double signs are taken, there must be at least 
one variation between 1 and 4, and one between 8 and 10, 
and there are evidently four between 4 and 8 ; or in all there 
are at least six variations in the result. As in the original 
equation there were five variations, the introduction of a 
positive root has caused at least one additional variation ; and 
as this is true of any positive root, there must be at least as 
many variations of sign as there are positive roots. 

Similarly, hy introducing the factor x + a, we may show 
that there are at least as many permanences of sign as there 
are negative roots. , 

If the equation is incomplete, any missing term must be 
supplied with ± as its coefficient before applying Descartes' 
Rule. 

508. In any complete equation, the sum of the number 
of permanences and variations is equal to the number of terms 
less one, or is equal to the degree of the equation (Art. 480). 
Hence, when the roots are all real, the number of positive 
roots is equal to the number of variations, and the number of 
negative roots is equal to the number of permanences (Art. 4S7). 



GENERAL THEORY OF EQUATIONS. 385 

A complete equation whose terms are all positive can have 
no positive root ; and one whose terms are alternately positive 
and negative can have no negative root. 

509. In an incomplete equation, imaginary roots may 
sometimes he discovered hy means of the douhle sign of in 
the missing terms. Thus, in the equation 

x 3 + x 2 ± x + 4 = 

if we take the upper sign, there is no variation, and conse- 
quently no positive root; if we take the lower sign, there is 
hut one permanence, and hence but one negative root. There- 
fore, as the equation has three roots (Art. 487), two of them 
must he imaginary. 

In general, whenever the term which precedes a missing 
term has the same sign as that which follows, the equation 
must have imaginary roots ; where it has the opposite sign, 
the equation may or may not have imaginary roots, hut 
Descartes' Hule does not detect them. If two or more suc- 
cessive terms of an equation he wanting, there must be imagi- 
nary roots. 

Note. In all applications of Descartes' Rule, the equation must con- 
tain a term independent of x, that is, no root must be equal to zero (Art. 
330) ; for a zero root cannot be considered as either positive or negative. 

EXAMPLES. 

510. The roots of the following equations being all real, 
determine their si cms : 



*&' 



1. x 8 -3x-2 = 0. 3. a; 8 -7 a; 2 + 36 = 0. 

2. .x 3 -10a; + 3 = 0. 4. x i -2x 3 -13x 2 + 38*-24 = 0. 

5. What are the signs of the roots of the equation x 3 + x 1 
-4 = 0? 

DERIVED POLYNOMIALS. 
511. If we take the polynomial 

a x n + b x n ~ 1 + c x n ~ 2 + 



386 ALGEBRA. 

and multiply each term by the exponent of a; in that term, 
and then diminish the exponent by 1, the result 

n a x" ~ l + (n — 1) b x H ~ 2 + Qi — 2) c x n ~ 3 + 

is called the first derived polynomial or first derivative of the 
given polynomial. 

The second derived polynomial or second derivative is the 
first derived polynomial of the first derivative ; and so on. 
The given polynomial is sometimes called the primitwe poly- 
nomial. 

A derived equation is one whose first member is a deriva- 
tive of the first member of another. 

1. Find the successive derivatives of x* + 5 x 1 + 3 x + 9. 

Result : First, 3 x 2 + 10 x + 3. 
Second, 6 x + 10. 
Third, 6. 
Fourth, 0. 

EXAMPLES. 

Find the successive derivatives of the following : 

2. a; 3 -5r + 6x-2. 4. a x* -bx % + ex -Sd. 

3. 2cc 2 -ic-7. 5. 7cc 4 -13a; 2 +8x-l. 

EQUAL ROOTS. 
512. Let the roots of the equation 

x n +px n -' 1 + qx n ~ 2 + + tx 2 + ux + v = (1) 

be a, b, c, Then (Art. 490), we have 

x n +2? x n ~ 1 + q x n ~ 2 + = (x — a) (x — b) (x — c) 

Putting x + y in place of x, 



(x + y) n + p (x + y)"- 1 + ... = (y + x -a) (y + x-b) ... (2) 
By Art. 399, the coefficient of y in the first member is 

nx n ~ 1 +p (n — I) x n ~ 2 + q (n — 2) x n ~* + (3) 



GENERAL THEORY OF EQUATIONS. 387 

which, we observe, is the first derivative of (1) ; and, as in 

Art. 491, regarding x — a, x — b, as single terms, the 

coefficient of y in the second member is 

(x — b) (x — c) (x — d) to n — 1 factors "1 

+ (x — a) (x — c) (x — d) to n — 1 factors ! 

+ (x — a) (x — b) (x — d) to ii — 1 factors | ^ ' 

+ ...... j 

As (2) is identical, by Art. 413 these coefficients are equal. 

Now if b = a, that is, if equation (1) has two roots equal to 
a, every term of (4) will be divisible by x — a, hence (3) will 
be divisible by the same factor ; therefore (Art. 486) the first 
derived equation of (1) will have one root equal to a. Sim- 
ilarly, if c = b = a, that is, if (1) has three roots equal to a, 
(3) will have two roots equal to a ; and so on. Or, in general, 

If an equation has n roots equal to a, its first derived equa- 
tion xv ill have n — 1 roots equal to a. 

513. From the principle demonstrated in Art. 512, it is 
evident that to determine the existence of equal roots in an 
equation we must 

Find the greatest common divisor of the first member and 
its first derivative. If there is no common divisor there can 
be no equal roots. If there is a greatest common divisor, by 
placing it equal to zero and solving the resulting equation we 
shall obtain the required roots. 

The number of times that each root is found in the given 
equation is one more than the number of times it is found 
in the equation formed from the greatest common divisor. 

If the first member of the given equation be divided by the 
greatest common divisor, the depressed equation will contain 
the remaining roots of the original equation. 

1. Find the roots of the equation 

x* - 14 x s + 61 x~ - 84 x + 36 = 0. 

Here the first derivative is 4 x s — 42 x % + 122 x — 84 ; the 
greatest common divisor of this and the given first member 



388 ALGEBRA. 

i s x 1 — 7 x + 6. Placing x 2 — 7 x + 6 == 0, we have, by the 
rules of quadratics, or by factoring, x = 1 or 6. Therefore 
the roots of the given equation are 1, 1, 6, and 6. 

EXAMPLES. 

Find all the roots of the following : 

2. x s-Sx-+13x-6 = 0. 4. .T 4 -6a; 2 -8a;-3 = 0. 

3. x 3 -7x 2 +16x-12 = Q. 5. cc 4 -24x 2 + 64z-48 = 0. 

514. When the equation formed from the greatest com- 
mon divisor is of too high a degree to be conveniently solved, 
we may in certain cases compare it with its own derived 
equation, and thus obtain a common divisor of a lower degree. 
Of course this can only be done when the equation formed 
from the greatest common divisor has equal roots. 

For example, required all the roots of 

x b - 13 x* + 67 x 3 - 171 x- + 216 x - 108 = 0. (1) 

Here the first derivative is 5 x* — 52 x 3 + 201 ,x 2 — 342 x 
+ 216 ; the greatest common divisor of this and the given 
first member is x 3 — 8 x 2 + 21 x — 18. We have then to solve 
the equation 

x 3 -8x 2 +21z-18 = 0. (2) 

The first derivative of (2) is 3 x 2 — 16 x + 21 ; the greatest 
common divisor of this and x 3 — 8 x 2 + 21 x — 18 is x — 3. 
Solving x — 3 = 0, we have x = 3; hence two of the roots of 
(2) are equal to 3. Dividing the first member of (2) by 
(x — 3) 2 or by x 2 — 6 x + 9, the depressed equation is 

x — 2 = 0, whence x = 2. 

Thus the three roots of (2) are 3, 3, and 2. Hence, the five 
roots of (1) are 3, 3, 3, 2, and 2. 

515. If an equation has two roots equal in magnitude, but 
opposite in sign, by changing the signs of the alternate terms 
beginning with the second we shall obtain an equation with 
these same two roots (Art. 498) ; then evidently the greatest 



GENERAL THEORY OF EQUATIONS. 389 

common divisor of the two first members placed equal to zero 
will determine the roots. 

For example, required all the roots of 

x* + 3 x 3 - 13 x--21 x + 36 = 0. (1) 

Changing the signs of the alternate terms, we have 

x 4 - 3 x 3 - 13 x°- + 27 x + 36, 

the greatest common divisor of which and the given first 
member is x 2 — 9 ; solving x 2 — 9 = 0, we have x = 3 or — 3. 
thus giving two of the roots of (1). Dividing the first mem- 
ber of (1) by x 2 — 9, we have for the depressed equation 

x- + 3 x - 4 = 0, 

whence x = 1 or — 4. Thus the roots of (1) are 3, — 3, 1, 
or —4. 

LIMITS OF THE ROOTS OF AN EQUATION. 
516. A polynomial of the form 

x n + p x n ~ * + q x n ~ 2 + J r tx 2 + ux+v 

which we shall represent by X, may also be expressed thus 
(Art. 490) : 

(x — a) (x — b) (x — c) (x — I) Y 

in which a, b, c, I are the real, unequal roots of the equa- 
tion X= 0, in the order of their magnitude, a being algebrai- 
cally the smallest; and Y the product of all the factors con- 
taining imaginary roots, which must always be positive, and 
cannot affect the sign of X, for each pair of imaginary roots 
(Art. 497) produces a positive factor. 

Suppose x to commence at any value less than a, and to 
assume in succession all possible values up to some quantity 
greater than I. When x is less than a, each of the factors 
x — a, x — b, is negative, and therefore X is either posi- 
tive or negative, according as the degree is even or odd. 



390 ALGEBRA. 

When x == a, X = 0. When x is greater than a, and less 
than b, x — a becomes positive, and the sign of X changes. 
Also, when the value of x is made equal to b, and then greater, 
X first becomes and then changes sign; and so on, for each 
real root. 

When x has any value greater, than I, X must be positive ; 
for all its factors are positive. 

517. If two numbers, when substituted for the unknown 
quantity in an equation, give results having a different sign, 
at least one root lies between those numbers. 

It is evident, from Art. 516, that if X has a different sign 
for two values of x, some odd number of roots lies between 
them. 

When the numbers substituted differ by unity, it is evident 
that the integral part of the root is known. 

EXAMPLES. 

1. What is the first figure of a root of the equation x 3 + 3 x- 

Here, if x = 2, the first member becomes — 2 ; and if x = 3, 
the first member becomes 25 ; therefore at least one root lies 
between 2 and 3. Hence 2 is the first figure of a root. 

2. Find the integral parts of all the roots of the equation 
x 3-Gx 2 +3x+9 = 0. 

3. Find the first figure of a root of the equation x 3 —2x 
- 50 = 0. 

4. Find the first figure of a root of the equation x* — 2 ar 8 
+ 3x 2 -x-5 = 0. 

5. Find the integral part of a root of the equation 2 x 4 + x : 
-7x 2 -llx-4, = 0. 

518. To find the superior limit of the positive roots of an 
equation. 

Let the equation be 

X= x n + p x"- 1 + q .r"- 2 + + tx- + ux + v = 0. (1) 



a 



GENERAL THEORY OF EQUATIONS. 301 

Let r be the numerical value of the greatest negative 
coefficient, and x n ~ s the highest power of x which has a nega- 
tive coefficient. Then the first s terms have positive coef- 
ficients. 

Now Xwill be positive when x is positive, provided 

x 11 — r x n ~ * — ) , / _,_1 — — r x- — r x — r (2) 

is positive ; for, since r is the numerically greatest negative 
coefficient, and all terms up to the (s + l)th are positive, Xis 
equal to (2) plus a, positive quantity. 
We may write (2) 

x n — r (x n - s + x n - s - 1 + + x °- + x+l), 

or (Art. 120), x*-r - — — ~ . (3) 

x — 1 

Then AT will be positive when (3) is positive. But if x is 
greater than unity, (3) is evidently greater than 

x n — s + 1 



X-l 



Therefore X will be positive when this is positive ; or, when 
(x — 1) x 11 — rx n - s + 1 is j>ositive ; or, when (x — 1) x s ~ x — r 
is positive. 

But (x— 1) X s-1 — r is greater than (x — 1) (x — l) s_1 — r 
or (x — 1)' — r ; therefore X will be positive when (x — l) s — r 
is positive or equal to zero ; or, when (x — l) s = r or > r ; 
or, when x — l — \Jr or > \/r; or, when x = l-\-\jr or >1 
+ </r. 

That is, when x = 1 + \J r or any greater value, X is posi- 
tive, which is impossible, as it must equal zero. Hence x 
must be less than l + tyr; or, 1 + $r is the superior limit of 
the positive roots. 

519. To find the inferior limit of the negative roots of an 
equation. 

By changing the signs of the alternate terms beginning 
with the second, we shall obtain an equation having the 
same roots with contrary signs (Art. 408). 



392 ALGEBRA. 

Then evidently the superior limit of the positive roots of 
the transformed equation, obtained as in Art. 518, will by a 
change of sign become the inferior limit of the negative roots 
of the given equation. 

Note. In applying the principles of the preceding articles to determine 
the limits of the roots of an equation, the absolute term must be taken as 
the coefficient of x'K 

520. 1. Find the superior limit of the positive roots of 

x 4 + 4 x 3 - 19 x- - 46 x + 120 = 0. 

Here, r = 46, and n — s = 2 ; or, as n = 4, s — 2. Then by 
Art. 518, the required limit is 1 + y/46, or 8 in whole num- 
bers. 

2. Find the inferior limit of the negative roots of 

a- 3 -a; 2 -14.x +24 = 0. (1) 

Changing the signs of the alternate terms beginning with 
the second, we have 

x 3 + x 2 -Ux-2A = 0. (2) 

Here r = 24, and n — s = 1, or s = 2. Then the superior 
limit of the positive roots of (2) is 1 + \/24 ; therefore the 
inferior limit of the negative roots of (1) is — (1 + v/24). 

EXAMPLES. 

Find the superior limits of the positive roots of the follow- 
ing: 

3. x * + 2x 3 -13.r' 1 -Ux + 24:=Q. 4. a: 4 -15.T 2 +10,r+24=0. 
Find the inferior limits of the negative roots of the fellow- 



s' 



mg 



5. x*-2x°--5x + (j = 0. 6. x x -5x 5 +ox- + 5x + 6 = 0. 

STURM'S THEOREM. 

521. To determine the number and situation of the real 
roots of an equation. 



GENERAL THEORY OF EQUATIONS. 393 

A perfect solution of this difficult problem was first obtained 
by Sturm, in 1829. As the theorem determines the number 
of real roots, the number of imaginary roots also becomes 
known (Art. 487). 

522. Let X denote the first member of 

X n +21X n - 1 + qx n ~ 2 + + tX 2 + UX+ V = 0, 

from which the equal roots have been removed (Art. 512). 

Let X x denote the first derivative of X (Art. 511). 

Divide AT by X 1} and we shall obtain a quotient Q l} with a 
remainder of a lower degree than X v Denote this remainder, 
with its signs changed, by X 2 , divide X x by X 2 , and so on ; 
the operation being the same as in finding the greatest com- 
mon divisor, except that the signs of every remainder must be 
changed, while no other change of signs is admissible. As 
the equation X — has been freed from equal roots, there can 
be no common divisor of Zand J 1; and the last remainder, 
X„ , will be independent of x. 

The successive operations may be represented by the fol- 
lowing equations : 

X = X x Q v - X 2 (1) 

X I = X 2 Q 2 -X 3 (2) 
X^X,Q,-X 4 (3) 



X n _ 2 — -<*« - 1 V« - 1 X n 

The expressions X, X 1} X 2 , X H are called Sturm's 

Functions. 

STURM'S THEOREM. 

523. If any tiro numbers, a and b, be substituted/or x in 
Sturm's Functions, and the signs noted, the difference between 
the number of variations in the first case and that in the 
second is equal to the number of real roots of the given equa- 
tion lying between a and b. 



394 ALGEBRA. 

The demonstration of Sturm's Theorem depends upon the 
following principles : 

(A). Two consecutive functions cannot both become for 
the same value of x. 

For, if A\ = and X = 0, then by (2), Art. 522, X 3 = ; 
and if X 2 = and X 3 = 0, by (3), X 4 = 0; and so on, till 
X n = 0. But as X n is independent of x, it cannot become 
for any value of x. Hence no two consecutive functions can 
become zero for the same value of x. 

(B). If any function, except X and X n , becomes for a 
■particular value of x, the two adjacent functions must have 
opjjosite signs. 

For, if X, = 0, we have by (2), Art. 522, X 1 = — X s ; that 
is, X 1 and X s must have opposite signs, for by (A) neither can 
be equal to zero. 

(C). When any function, except X and X n , changes its sign 
for different values of x, the number of variations is not 
affected. 

No change of sign can take place in any one of Sturm's 
Functions except when x passes through a value which re- 
duces that function to zero. 

Now, let c be a root of the equation X 2 = 0; d and e quan- 
tities respectively a little less and a little greater than c, so 
taken that no root of X x = or of X s = is comprised be- 
tween them. Then, as x changes from d to e, no change of 
sign takes place in A^ or X :i , while X 2 reduces to zero and 
may change sign. And as by (B), when X 2 = 0, X l and X z 
have opposite signs, the only effect of a change in the sign of 
X 2 is that what was originally a permanence and a variation 
is now a variation and a permanence ; that is, the permanence 
and variation exchange places. Hence a change in the sign 
of X., does not affect the number of variations. 

As X n is independent of x, it can never change sign for any 
value of x. Therefore a change in the number of variations 



GENERAL THEORY OF EQUATIONS. 395 

can be caused only by a change in the sign of the given 
function X. 

(D). When the function X changes its sign for successive 
increasing values of x, the number of variations is diminished 
by one. 

Let m be a root of the equation X = ; m — y and m + y 
quantities respectively a little less and a little greater than 
7>i, so taken that no root of X 1 = Q is comprised between them. 
Then, as x changes from m — y to m + y, no change of sign 
takes place in X r , while X reduces to zero and changes sign. 

Putting in + y in place of x in X, we have 

O + y) n + P (m + y) n ~ l + + u (m + y) + v. 

B/eveloping the terms by the Binomial Theorem, and col- 
lecting terms containing like powers of y, we have 

m n + p m n ~ 1 + + um + v 

+ y \_n m" _ l + p (n — 1) ni n ~~ 2 + + u\ 

+ terms containing y 2 , y s , y n . 

Representing the coefficient of y, which we observe is the 
value of X 1 when x is put equal to m, by A ; the coefficient of 
y 1 by B; and so on, we have 

m n +pm n - 1 + + um + v + Ay + By-+ + Ky n . (1) 

But as x = m reduces X to 0, we have 

m n + p m n ~ 1 + + u m + v = 0. 

Hence (1) may be written 

Ay + Bf-+ + Ky\ (2) 

Now y may be taken so small that the sign of (2) will be 
the same as the sign of its first term. That is, when a? is a 
little greater than m, the sign of X is the same as the sign 
of X l . 

Similarly, by substituting in — y for x in X, we shall arrive 
at the expression 

-Ay+By 2 -Cy 3 + , 



396 ALGEBRA. 

where as before y may be taken so small that the sign of the 
whole expression will be the same as that of its first term. 
That is. when a; is a little less than m, the sign of X is the 
reverse of the sign of X x . 

Thus we see that as x changes from m — y to m + y, the 
signs of X and A^ are different before x equals m, and alike 
afterwards. Hence, when A changes its sign a variation is 
changed into a permanence, or the number of variations is 
diminished by one. 

We may now prove Sturm's Theorem ; for as x changes 
from a to b, supposing a less than b, a variation is changed to 
a permanence each time that X reduces to and changes sign, 
and only then, for no change of sign in any of the other 
functions can affect the number of variations. And as X 
reduces to zero only when x is equal to some root of the 
equation X — 0, it follows that the number of variations lost 
in passing from a to l> is equal to the number of real roots of 
the equation X = comprised between a and b. 

524. When — oo and + go are substituted for x, or when 
the superior limit of the positive roots and the inferior 
limit of the negative roots are substituted for x, the whole 
number of real roots of the equation A^ = becomes known. 

The substitution of — go and will give the whole number 
of negative roots, and the substitution of + go and will give 
the whole number of positive roots. If the roots are all real, 
Descartes' Kule (Art. 507) will effect the same object. 

The substitution of various numbers for x will show be- 
tween what numbers the roots lie, or fix the limits of the 
roots. 

525. A and A, must change signs alternately, as they are 
always unlike in sign just before X changes sign (Art. f>L\' > > i 
(D)). Hence, when the roots of X = and of Aq = are all 
real, each root of Aq = must be intermediate in value be- 
tween two roots of X= 0. For this reason the first derived 
equation is often called the limiting or separating equa- 
tion. 



GENERAL THEORY OF EQUATIONS. 397 

526. In the process of finding X 2 , X 3} etc., any positive 
numerical factors may be omitted or introduced at pleasure, as 
the sign of the result is not affected thereby. In this way 
fractions may be avoided. 

In substituting — go and + go, the first term of each func- 
tion determines the sign, for in any expression, as 

ax n + bx n ~ 1 + + /,•, 

where x may be made as great as we please, it may be taken 
so great that the sign of the whole expression will be the same 
as that of its first term. 

527. 1. Determine the number and situation of the real 
roots of the equation 

x 3 — 4 x 2 — x + 4 = 0. 

Here, the first derivative, X 1 = 3 x 2 — 8 x — 1. Multiply- 
ing x 3 — 4 x' 1 — x + 4 by 3 so as to make its first term divisi- 
ble by 3 x 2 , 

3x 2 -8x-l)3x 3 -12x 2 - 3 a: + 12 (a; 



3 a; 3 - Sx 2 



x 



- 4r- 2 a; + 12 

3 

- 6 a; 2 - 3 x + 18 (-2 

- 6.r- + 16a- + 2 



19 a: + 16 .-. X = 19 x - 16. 



3 x 2 - 8 x - 1 
19 

19 x - 16 ) 57 x 2 - 152 x - 19(3x 

57 x 2 — 48 x 

- 104 x - 19 
19 



1976 a;- 361 (-104 
1976 x + 1664 

-2025 .-. X = 2025. 



398 ALGEBRA. 

Thus we have, X = x z — 4 x 2 — x + 4 ; X 2 = 19 cc — 10. 
Xj == 3 x- - 8 x - 1 ; Z 3 = 2025. 

The last step of the division may he omitted, for we only- 
wish the sign of X 3 , and that may he seen by inspection when 
— 104 x — 19 is obtained. 

We first substitute — go for x in each function, and obtain 
three variations of sign ; similarly + go gives no variation ; 
hence the three roots are all real. Substituting 0, we have 
two variations ; comparing this with the former results, we 
see that one root is negative and the other two are positive. 
The same result could have been obtained by Descartes' Rule, 
as all the roots are real. We now substitute various numbers 
to determine the limits of the roots. 

The table presents the results in a connected form : 







X 


*1 


^2 


*3 




When 


X = — 00, 


— 


+ 


— 


+ 


3 variations. 


Li 


x = — 2, 


— 


+ 


— 


+ 


3 variations. 


a 


x = — 1, 





+ 


— 


+ 




it 


x = 0, 


+ 


— 


— 


+ 


2 variations. 


n 


X = 1, 





— 


+ 


+ 




a 


x = 2, 


— 


— 


+ 


+ 


1 variation. 


a 


% == O} 


— 


+ 


+ 


+ 


1 variation. 


u 


x = 4, 





+ 


+ 


+ 




a 


OC — Oj 


+ 


+ 


+ 


+ 


no variation. 


a 


X = GO, 


+ 


+ 


+ 


+ 


no variation. 



Then by Sturm's Theorem we know that there is one root 
between —2 and 0, one between and 2, and one bet ween 3 
and 5. In fact, as X == when x = — 1, 1, and 4, these are 
the three roots of the equation. 

2. Determine the number and situation of the real roots of 



X 



x 4 - 3 x s + 3 x- — 3 x + - = 



Note. In substituting the various numbers to determine the situation 
of the roots, it is best to work from o in cither direction, stopping when 
the number of variations is the same as has been previously found for 
+ oo or — oo , as the case may be. 



SOLUTION OF HIGHER NUMERICAL EQUATIONS. 399 

Here we find X, = 4x 3 - 9x 2 + 6x - 3 ; X a = — <d2x + 129 ; 
X = 3 x 2 + 18 x - 31 ; X i = - 1163. 

Substituting + co for x, we obtain one variation ; similarly, 
gives three variations, and — oo gives three variations. 
Hence there are only two real roots, both of which are posi- 
tive. We then substitute values of x from upwards, giving 
the following results : 





X 


^ 


x 2 


X s X 4 




When x = 0, 


+ 


— 


— 


+ — 


3 variations. 


■ " x = l, 


+ 


— 


— 


+ - 


3 variations. 


" x = 2, 


+ 


+ 


+ 


— — 


1 variation. 


" x = co, 


+ 


+ 


+ 


— — 


1 variation. 



Hence there are two roots between 1 and 2 ; and as the 
equation has four roots, there must be two imaginary roots. 

EXAMPLES. 

Determine the number and situation of the real roots of the 
following equations : 

3. x*-x 2 -2x + l = Q. 6. x*-2x 9 -5x 2 + 10a;-3 = 0. 

4. g* — 7x+7 = 0. 7. 2x- 4 -3* 3 + 17x- 2 -3x- + 15 = 0. 

5. X s — 2 a -5 = 0. 8. x*- 4 X s -3 x + 27 = 0. 



XLIIL — SOLUTION OF HIGHER NUMERICAL 

EQUATIONS. 

528. The real roots of the higher numerical equations in 
general can only be obtained by tentative methods, or by 
methods which involve approximation. Cubic and biquadratic 
equations may be considered as included in the class of higher 



400 ALGEBEA. 

equations ; for their general solutions are complicated, and 
only of limited application. No general solution of an equa- 
tion of a degree higher than the fourth can he ohtained. 



COMMENSURABLE ROOTS. 

529. A commensurable root is one which can be exactly 
expressed as an integer or fraction without using irrational 
quantities. 

An incommensurable root is one which can only he ex- 
pressed approximately by means of a decimal fraction. 

530. Any equation containing fractional coefficients may 
be transformed into another whose coefficients are entire, that 
of the first term being unity (Art. 500), and such an equation 
cannot have a root equal to a rational fraction (Art. 495) ; 
hence, to find all commensurable roots, we have only to find 
all integral roots. 

531. As every rational root of an equation in its general 
form is a divisor of the last term (Art. 493), to find the com- 
mensurable roots we have only to ascertain by trial what in- 
tegral divisors of the absolute term are roots of the equation. 

The trial may be made by substituting each divisor, both 
with the positive and the negative sign, in the equation ; or 
by dividing the first member of the equation by the unknown 
quantity minus the supposed root (Art. 486). In substi- 
tuting very small numbers, such as ± 1, the former method 
may be most convenient ; but when an actual root lias once 
been used, the latter method will give at once the depressed 
equation, which may be used in obtaining the other roots. 

532. When the number of divisors of the last term is 
large, this process of successive trials becomes tedious, and a 
better method, known as the Method of Divisors, may be 
adopted. 

If a is a root of the equation 

x 4 +px i +qx 2 + tx + u = 0, 



SOLUTION OF HIGHER NUMERICAL EQUATIONS. 401 

then a 4 + p a s + q a- + t a + u= 0. 

Transposing and dividing by a, 

-= — t — qa—pa 2 — a 3 , (1) 

u 
whence we see that - must be an integer. 

a 

Equation (1) may be written 

— h t = — q a —p a- — a*, 
cv 



U 

a 

a 



u 
Denoting - + t by t', and dividing by a, 

— — q —p a — a 2 , 



t' 
whence - must be an integer. 

a 



Proceeding in this way, we see that if a is a root of the 
equation, — \- t or t', — \- q or q', and \- p or p 1 must be in- 

Ct Ct Lt 

p' 

tegers, and 1-1 must equal zero. 

Hence the following 

RULE. 

Divide the absolute term of the equation by one of its inte- 
gral divisors, ami to the quotient add the coefficient of x. 

Divide this sum by the same divisor, and, if the quotient is 
an integer, add, to it the coefficient ofx 2 . 

Proceed in the same manner with each coefficient in regular 
order, and, if the divisor is a root of the equation, each 
quotient will be entire, and the last quotient added to the 
coefficient of the highest poicer ofx will equal 0. 

Equal roots, if any, should be removed before applying the 
rule ; and the labor may often be diminished by obtaining the 
superior limit to the positive and inferior limit to the nega- 



402 ALGEBRA. 

tive roots of the equation, for no number need be tried which 
does not fall between these limits. 

1. Find the roots of the equation 

x s - 6 x 2 + 27 x - 38 = 0. 

By Descartes' Rule, we see that the equation has no nega- 
tive root; and the only positive divisors of 38 are 1, 2, 19, 
and 38. By substitution we see that 1 is not a root of the 
equation. 

Dividing the first member by x — 2, we obtain x 2 — 4 x + 19 
as a quotient. Hence 2 is a root, and the depressed equation 
is x 2 — 4 x + 19 = 0, from which we obtain 



x = 



4±y/16-76 

~~2~ 

as the remaining roots. Hence, 



2±v/-15 



x = 2, or 2 ± y/ — 15, Ans. 
2. Find the roots of the equation 

8 x* - 4 x s - 14 x 2 + x + 3 = 0. 
We may write the equation 

A 9C i QC 0C O - 

*-2--^r + § + 8= - 

Proceeding as in Art. 500, we see that the multiplier 2 will 
remove the fractional coefficients. We then have the equation 

x 4 — x s — 7 x 2 + x + 6 = 0, (1) 

whose roots are twice those of the given equation (Art. 499). 

The divisors of G are ±1, ±2, ±3, and ± 6. 

By putting x equal to + 1 and — 1 in (1), it is readily seen 
that both are roots of the equation, and the other roots can be 
found from the depressed equation. But all of the rational 
roots may be obtained by the rule. 



6, 


3, 


2, 


1, 


-1, 


-2, 


-3,-6 


1, 


2, 


3, 


6, 


-6, 


-3, 


-2, -1 


2, 


3, 


4, 


("7 


-5, 


-2, 


-1, 




1, 


2, 


7, 


5, 


1, 







-6, 


-$ 


o, 


9 


-6, 


-7 



-2, 


o, 


2, 


3 


-3, 


-1, 


1, 


2 


-1, 


-1, 


-1, 


-1 


o, 


o, 


o, 






SOLUTION OF HIGHER NUMERICAL EQUATIONS. 403 

It is customary to abridge the work as follows : 

Divisors, 
1st Quotients, 
Adding 1, 
2d Quotients, 
Adding — 7, 
3d Quotients, 
Adding — 1, 
4th Quotients, 
Adding 1, 

As 6, 2, — 3, and — 6 give fractional quotients at different 
stages of the operation, they cannot be roots of the given 
equation, and are rejected. 3, 1, — 1, and — 2 give entire 
quotients, and in each case the last quotient added to the 
coefficient of x 4 gives zero ; hence they are the four roots of 

3 1 1 

equation (1), and ^, ^, — ^, and — 1 are the four roots of the 

given equation. 

EXAMPLES. 

Find all the commensurable roots of the following equa- 
tions, and the remaining roots when possible by methods 
already given : 

3. x s +Qx 2 +llx + Q = 0. 8. x s -7x 2 + 36 = 0. 

4. x s + 3x 2 -4;x-12 = 0. 9. x 3 -6x 2 + 10a; — 8 = 0. 

5. a- 4 -4.r 3 -8x + 32 = 0. 10. x 3 - 6 x 2 + 11 x- 6 = 0. 

6. 4a; 8 -16sc 2 -9a; + 36 = 0. 11, 2x 3 -3x 2 +16x-2i = 0. 

7. x 3 -3x 2 + x + 2 = 0. 12. x 5 - 2 x 3 -16 = 0. 

13. x i -9x 3 + 23x 2 -20x + 15 = 0. 

14. x 4 + x 3 - 29 x 2 - 9 x + 180 = 0. 



404 ALGEBRA. 



RECURRING OR RECIPROCAL EQUATIONS. 

533. A Recurring Equation is one in which the coef- 
ficients of any two terms equally distant from the extremes 
of the first member are equal. 

The equal coefficients may have the same sign, or opposite 
signs; hut a part cannot have the same sign, and a part 
opposite signs, in the same equation. Also, if the degree he 
even, and the equal coefficients have opposite signs, the middle 
term must be wanting. Thus, 

a 4 - 5 x z + 6 x 2 - 5 x + 1 = 0, 

5 x 5 - 51 x* + 160 x 3 - 160 x 2 + 51 x - 5 = 0, 

x 6 — x 5 + x* — x 2 + x — 1 = 0, 

are recurring equations. 

534. If any quantity is a root of a recurring equation, 
the reciprocal of that quantity is also a root of the same 
equation. 

Let x n +px n ~ 1 + qx n ~ 2 + ...± (... + </x 2 +i*z + l)=0 (1) 

be the equation. Substitute - for x ; then 

V 

y" y'" 1 y n ~- v v v > 

Multiplying each term by y n , 

(i+py+.qf.+,..)±(...+ qy n -*+py n - 1 +y n )=Q ( 2 ) 

Now, (1) and (2) take precisely the same form on changing 

the ± sign to the first parenthesis in equation (2), and hence 

they must have the same roots. Now, if a is a root of (1), as 

1 1 
v — - must be a root of (2) ; but, as (1) and (2) have the 
x a 

same roots, - must also be a root of (1). In like manner, if 
a 

b is a root of (1), y is also a root of (1). 



SOLUTION OF HIGHER NUMERICAL EQUATIONS. 405 

On account of the property just demonstrated, recurring 
equations are also called reciprocal equations ; the former term 
relating to their coefficients, and the latter to their roots. 

535. One root of a recurring equation of an odd degree is 
— 1 when the equal coefficients have the same sign, and +1 
when they have opposite signs. 

A recurring equation of an odd degree, as 

x *n + l +px 2m + qx 2m-l + _ ± ( + q tf + p x + ^ _ Q (3) 

has an even number of terms, and may he written in one of 
the following forms, 

(x 2M + 1 + 1) +p (x 2m + x) + q (x 2 "'- 1 + x") + =0, 

( X 2m + 1 — 1) +p (x 2m — x) + q (X lm - 1 -X-) + ...... = 0. 

If — 1 he substituted for x in the first form, or + 1 in the 
second, the first member will become ; hence, — 1 is a root 
of the first and + 1 a root of the second. 

If equation (3) be divided by x ± 1, both forms will reduce 
to the following form, 

X 2m +px 2 " 1 - 1 + qx 2m ~ 2 + + qx 2 +px + l = 0, (4) 

a recurring equation of an even degree in which the equal 
coefficients have the same sign. Hence, a recurring equation 
of an odd degree may always be depressed to one of an even 
degree. 

536. Two roots of a recurring equation of an even degree 
are + 1 and — 1 when the equal coefficients have opposite signs. 

Let 

x *>» + pX 2m - 1 + qX 2m - 2 + -( + qx 2 +px+ 1)=0 

be such an equation. As the middle term must be wanting 
(Art. 533), the equation may be written in the form 

^™-l)+ F (^»" ! -l) + ?x 2 (f'»- 4 -l)-h = (5) 

which is divisible by both x — 1 and x + 1, or by x' 2 — 1 (Art. 
120). Hence, both + 1 and — 1 are roots of the equation. 



406 ALGEBRA. 

If equation (5) be divided by x 1 — 1, it will be depressed 
two degrees, and become a recurring equation of an even 
degree, in which the equal coefficients have the same sign 
(Art. 120). Hence, every recurring equation may be de- 
pressed to the form of equation (4), Art. 535. 

537. Every recurring equation of an even degree, whose 
equal coefficients have the same sign, may be reduced to an 
equation of half that degree. 

Let 

a? m +qi x-"'- 1 + q ar m ~ 2 + + q x~ +p x + 1 = 

be such an equation. Dividing it by x m , we may write it 

(^ + ^) + ^(^"- , + ^)+?(^- 2 + ^r- 2 ) + ---=0 (6) 

the middle term if present becoming a known quantity. 
Put x + - = y 

Then, a 2 + -=- = y 2 - 2 

x- 

x3+ x^ =y3 ~ s ( x+ x) =i/3 ~ s y 



x m + — = y m — m y m ~ 2 + 

Substituting these values in (6), we have an equation of the 
form 

y" l +2hy m - l + qiy m -* + = 0. 

After this equation is solved, we can immediately find x 

from the equation x + - = ?/. 

x 

538. It thus appears that any recurring equation of the 
(2 m + l)th degree, one of the (2 m + 2)th degree whose equal 



SOLUTION OF HIGHER NUMERICAL EQUATIONS. 407 

coefficients have opposite signs, and one of the 2 with degree 
whose equal coefficients have the same sign, may each he 
reduced to an equation of the with degree. 

EXAMPLES. 

1. Given x 4 — 5 x s + 6 x 2 — 5 x + 1 = 0, to find x. 
Dividing by x 2 , far -\ — j j — 5 f x -\ — j + 6 = 0. 

S instituting y for x -\ — , and y 2 — 2 for x 2 -\ — - 9 we have 

x x l 

w 2 -2-5w+6 = 0. 

Whence, y = 4 or 1. 

If y = 4, x + - = 4, or x 2 — 4 x = — 1 ; 

Whence, x = 2 ± ^3. 

If y = 1, x + - = 1, or a; 2 — x = — 1 ; 



Whence, x = T 

Note. That 2 — y/3 and are reciprocals of 2 + y/3 and - — ^ 

1 2 

may easily be shown by reducing — — — and — - — ;=. to equivalent frac- 

2 + ^3 1 + V— 3 

tions with rational denominators (Art. 279). 
Solve the following equations : 

2. a- 5 - 11 x l + 17 x 3 + 17 a- 2 - 11 x + 1 = 0. 

3. a; 5 + 2 x* - 3 a; 3 - 3 x 2 + 2 x + 1 = 0. 

4. x G — x 5 + x 4 — x 2 + x — 1 = 0. 

5. a; 3 + px 2 +px + 1 = 0. 

6. 6 x 4 + 5 a.- 3 - 38 x 2 + 5 x + 6 = 0. 

7. 5 x 5 - 51 a,- 4 + 160 a; 3 - 160 a- 2 + 51 x - 5 = 0. 



408 ALGEBRA. 

8. x 4 + 5 x' + 5 x + 1 — 0. 

9. x 5 = - 1, or x> + 1 = 0. (See Art. 332.) 
10. x 5 -32 = 0. (Let cc = 2 ?/.) 

CARDAN'S METHOD FOR THE SOLUTION OF CUBIC 

EQUATIONS. 

539. In order to solve a cubic equation by Cardan's 
method, it must first be transformed, if necessary, into another 
cubic equation 1 in which the square of the unknown quantity 
shall be wanting. 

By Art. 505, this may be done by substituting for x, y 
minus the coefficient of x 2 divided by 3. 

540. If the first power of the unknown quantity be want- 
ing in the given equation, we may obtain the result by a 
simpler method, as follows : 

Let x 3 + a x 2 + c = be such an equation. 

Substituting - for x, we have 
V 

1 a 

— + — + c = 0, or c y % + a y + 1 = 0. 

541. To solve a cubic equation in the form x 3 + p x + q = 0. 

P 
Put x = z — =-, and the equation becomes 
Sz 

V 2 V 3 7) 2 

or, z 3 - £—. + q = 0; or, 27 z* + 27 q z 3 -p 3 = 0. 

2( z 3 

This is an equation in the quadratic form, and may be 

solved by the method of Art. 313 ; and after z is known, x 

P 
may be found directly from the equation x = « — —-. 

o z 



SOLUTION OF HIGHER NUMERICAL EQUATIONS. 409 
We have then for solving cubic equations the following 

RULE. 

If necessary, transform the equation into another cubic 
equation in which the square of the unknown quantity shall 
be wanting (Arts. 539 and 540). 

If y be the unknown quantity in the resulting equation, 
substitute for it z minus the coefficient of y divided by 3z. 

EXAMPLES. 

1. Solve the equation x 3 — 9 x + 28 = 0. 

3 

Substituting z -\ — for x, 

27 97 27 

z 3 + 9z + — +^--9z - — + 28 = 0, 

z z 6 z 

27 
or, « 3 + tf + 28 = 0; or, * c + 28s 3 = -27. 

Solving by quadratics, z 3 = — 1 or — 27. 
Whence, z = — 1 or — 3. 

Uz = -1, x = z + - = -l-3 = -±. 

z 

If z = -3, x = - 3 - 1 = - 4. 

Hence, one root of the equation is — 4. Dividing the first 
member of the given equation by x + 4, we obtain as the de- 
pressed equation, 

x 2 — 4 x + 7 = 0. 



Whence, x — 2 ± y/— 3, the remaining roots. 

2. Solve the equation x s — 24 x 2 — 24 x — 25 = 0. 
Putting x = y + 8 (Art. 539), we obtain 
2/ 3 + 24^+192 Z / + 512-24y 2 -384?/-1536-24y-192-25=0, 
or, y 8 - 216 y — 1241 = 0. 



410 ALGEBRA. 

72 
Putting y = z H — -, we have 

. _,_ 15552 373248 ._ 15552 , OM _ 

z z + 216s h 1 5 216 « 1241 = 0, 

z z* z 

or, z 3 + 3 ' 3 ^ 48 - 1241 = ; or, s c - 1241 z 3 + 373248 = 0. 

Whence, s s = 729 or 512, and 2 = 9 or 8. 

72 72 

Therefore, y = 9 + - r or 8 + -^ = 17, and sc = y + 8 = 25. 

9 o 

Hence, one root of the equation is 25. Dividing the first 
member of the given equation by x — 25, we have as the 
depressed equation 

x 2 + x + 1 = 0: 



- 1 ± \J- 3 , 

Whence, x = - , the remaining roots. 

ii 

Solve the following equations : 

3. x 3 - 6x + 9 = 0. 6. x 3 + 9 x 2 - 21 x + 11 = 0. 

4. x 3 - 6 x 2 + 57 x -196=0. 7. x 3 -2 x 2 + 2x- 1 = 0. 

5. :c 3 -4a; 2 -3x + 18 = 0. 8. x 3 -±x 2 + 4a;- 3 = 0. 

9. a- 3 -3x 2 + 4 = 0. 
10. Obtain one root of the equation x 3 + 6 x — 2 = 0. 

542. In the cubic equation x 3 + px+q = 0, when p is 

— p 3 o 2 
negative, and f, > =y , Cardan's method involves imaginary 

expressions ; but it may be shown in that case that the three 
roots of the equation are then real and unequal. 

Thus, in solving the equation x 3 — 6 x + 4 = 0. 

2 

Substituting z + - for x, we have 
z 

z 3 + 6 2 H \- -j — Gz f- 4 = 0, 



SOLUTION OF HIGHER NUMERICAL EQUATIONS. 411 
r, g« + J* + 4 = ; or, z r > + 4 z* + 8 = 0. 



Whence, z z = -2 ± y/- 4, or -2 ± 2 y/- 1, 



or, ft = y/_ 2 + 2y/- 1 or ^-2-2^-1. 

It may be proved by trial that 1 + y/— 1 is the cube root of 
_ 2 + 2 yC3, and 1 - y/^1 of - 2 - 2 yCl. Hence, 



= l + y/-l or l-y/-l. 



If z = 1 + y/- 1, 



X=Z+-=1+ y/— 1 + 



2 _2y/-l + 2 

i + ^/=3 _ ' 1 + y'-i 



_ o 



Hence, one root of the equation is 2. Dividing the first 
member of the given equation by x — 2, we have as the de- 
pressed equation 

x 2 + 2 x - 2 = 0. 

Whence, £C = — 1 ± y^, the remaining roots. 

543 We have no general rule for the extraction of the 
cube root of a binomial surd ; so that in examples like that 
in the preceding article, unless the value of z can be obtained 
by inspection, it is impossible to find the real values of x by 
Cardan's method. . In this case, the real values of x can 
always be found by a method involving Trigonometry. 

BIQUADRATIC EQUATIONS. 

544. General solutions of biquadratic equations have been 
obtained by Descartes, Simpson, Euler, and others. Some of 
them require the second term of the equation to be removed, 
while others do not. All of them depend upon the solution of 
a cubic equation by Cardan's method, and will of course fail 
when that fails (Art. 542). They are practically of little 
value, especially as numerical equations of all degrees can be 
readily solved by methods of approximation. 



412 ALGEBRA. 

INCOMMENSURABLE ROOTS. 

545. If a higher numerical equation is found to contain 
no commensurable roots, or if, after removing the commen- 
surable roots, the depressed equation is still of a higher 
decree, the irrational or incommensurable roots must next be 
sought. The integral parts of these roots may be found by 
Sturm's Theorem or by Art. 517, and the decimal parts by 
any one of the three following methods of approximation. 

HORNER'S METHOD. 

546. Suppose a root of the equation 

x" +j>x"~ 1 + q x n ~ 2 + + tx' 2 + ux + v = (1) 

is found to lie between a and a + 1. Transform the equation 
into another whose roots shall be less by a (Art. 502), and we 
shall have an equation in the form 

y n +p'V n ~ 1 + Q'y n ~ 2 + + t'y°- + u'y + v' = 0, (2) 

one of whose roots is less than 1. If that root is found to lie 
between the decimal fractions a' tenths and a' + 1 tenths, 
transform equation (2) into another whose roots shall be less 
by a' tenths, and we shall have an equation in the form 

s»+_p"2"- 1 + q"z n ~ 2 + ...... + t"z 2 + u"z + v" = (3) 

one of whose roots is less than .1. If that root is found to 

lie between the decimal fractions a" hundredths and a" + 1 

hundredths, transform equation (3) into another whose roots 

shall be less by a" hundredths ; and so on. 

Thus we obtain 

x = a + a' + a" + 

to any desired degree of accuracy. 

As y and z in equations (2) and (3) are fractional, their 
higher powers are comparatively small ; hence approximate 
values of y and z may be found by considering the last two 
terms only, from which we have 

v' .. v" 

V = -77 and * ^ ~ TJ, ' 



SOLUTION OF HIGHER NUMERICAL EQUATIONS. 413 

Thus approximate values of a', a", may be found in 

this way, and with greater accuracy the smaller they become. 

Hence a positive incommensurable root of the equation may 
be found by the following 

RULE. 

Find by Sturm's Theorem the initial part of 'the root, and 
transform the given, equation into one whose roots are less by 
th is initial part. 

Divide the absolute term of the transformed equation by 
the coefficient of the first power of the unknown quantity for 
the next figure of the root. 

Transform this last equation into another whose roots are 
less by the figure of the root last found, divide as before for 
the next figure of the root ; and so on. 

547. A negative root may be found by changing the 
signs of the alternate terms of the equation beginning with 
the second, and finding the corresponding positive root of the 
transformed equation (Art. 498). This hy a change of sign 
becomes the-required negative root. 

548. In obtaining the approximate value of any one of 

the quantities a 1 , a", by tbe rule, we are liable to get too 

great a result ; a similar case occurs in extracting the square 
or cube root of a number. We may discover such an error 
by observing the signs of the last two terms of the next 
transformed equation ; for, as the figures of the root as ob- 
tained in succession are to be added, it follows that a', a", 

must be positive quantities, so that the last two terms of the 
transformed equation must be of opposite sign. We then 
diminish the approximate value until a result is found which 
satisfies this condition. 

549. If in any transformed equation the coefficient of the 
first power of the unknown quantity should be zero, the next 
figure of the root may be obtained by dividing the absolute 
term by the coefficient of the square of the unknown quantity, 
and taking the square root of the result. 



414 ALGEBRA. 

For, if in equation (2), Art. 546, u' = 0, we have, approxi- 
mately, 

t' y 2 + v' = 0, whence y = \/ 

We proceed in a similar manner if any numher of the 
coefficients immediately preceding the absolute term reduce 
to zero. 

550. 1. Solve the equation x 3 — 3 x 2 — 2 x + 5 — 0. 

By Sturm's Theorem, the equation has three real roots ; 
one between 3 and 4, another between 1 and 2, the third 
between — 1 and — 2. 

To find the first root, we transform the equation into 
another whose roots are less by 3, which by Art. 503 is 
effected as follows : 

Dividing x 3 — 3 x 2 — 2 x + 5 by x — 3, we have x 2 — 2 as a 
quotient and — 1 as a remainder. Dividing x 2 — 2 by x — 3, 
we have x + 3 as a quotient and 7 as a remainder. Dividing 
x + 3 by x — 3, we have 1 as a quotient and 6 as a remain- 
der. Hence the transformed equation is 

x 3 + 6 x 2 + 7 x - 1 = 0, 

whose roots are less by 3 than those of the given equation. 

Note. The operations of division in Horner's Method are usually per- 
formed by a method known as Synthetic Division. For example, let it be 
required to divide ofi - 19 x + 30 by x - 2. 



* 3 ±0.r 2 -19a; + 30 

T 3 _ O r 2 



x -2 



a; 2 + 2x-15 



2 a; 2 
2.-C 2 - ix 



-15a; 
-15.Z + 30 




The first term of each partial product may be omitted, as it is merely a 
repetition of the term immediately above. Also the remaining term of 
each partial product may be added to the corresponding term of the divi- 
dend, provided we change the sign of the second term of the divisor before 



SOLUTION OF HIGHER NUMERICAL EQUATIONS. 415 

multiplying. Also the powers of x may be omitted, as we need only 
consider the coefficients in order to obtain the remainder. 
The work now stands 

l±0-19 + 30|l+2 



+ 2 | 1 + 2-15 

+ 2 
+ 4 
-15 

-30 


As the first term of the divisor is 1, it is usually omitted, and the first 
terms of the dividends constitute the quotient. Raising the oblique 
columns we have the following concise form : 

Dividend, l±0-19 + 30|+2 

Partial Products, +2+ 4-30 

Quotient, 1 + 2-15,+ Remainder. 

Here we use only the second term of the divisor with its sign changed ; 
each term of the quotient is the sum of the terms in the vertical column 
under which it stands, and each term of the second line is obtained by 
multiplying the preceding term of the quotient by the divisor as written. 

By the method of Synthetic Division, the work of trans- 
forming the given equation into one whose roots are less by 
3 stands as follows : 



1 - 


3 


— 


2 


+ 5| +3 


+ 


3 







- 6 







— 


2 


— 1, 1st Remainder 


+ 


3 


+ 


9 




+ 


3 


+ 


7, 


2d Remainder. 


+ 


3 









-f 6, 3d Remainder. 

Thus the transformed equation is, as before, 

x 3 + 6 x 2 + 7 x - 1 = 0. (1) 

Dividing 1 by 7 we obtain .1 as the next figure of the root, 
and we proceed to transform equation (1) into another whose 
roots shall be less by .1. 



416 





ALGEBRA. 




6 


7 


-1 


.1 


.61 


.761 


6.1 


7.61 


-.239 


.1 


.62 




6.2 


8.23 




.1 







6.3 

Thus the transformed equation is 

x s + 6.3 x 2 + 8.23 x - .239 = 0, 
whence by dividing .239 by 8.23 we obtain .02 for the next root 
figure ; and so on. Thus the first root is, approximately, 3.12. 

Similarly, the second root may be shown to be 1.201 ap- 
proximately. 

By Art. 547, the third root is the positive root of the 
equation x 3 + 3 x~ — 2 x — 5 — with its sign changed. The 
successive transformations are usually written in connection 
as in the following form, where the coefficients of the different 

transformed equations are indicated by (1), (2), (3), 

The work may also be contracted by dropping such decimal 
figures from the right of each column as are not needed for 
the required degree of accuracy. 

1 



3 


_ 9 

4 


-5 | 1.33 


1 


2 


4 


2 


(1)173 


1 


5 


2.667 


5 


(1)7 


(2)- .333 


1 


1.89 




(1)6 


8.89 




.3 


1.98 




6.3 


(2) 10.87 




.3 






6.6 






.3 







(2) 6.9 
Hence, the third root is — 1.33 approximately. 



SOLUTION OF HIGHER NUMERICAL EQUATIONS. 417 

EXAMPLES. 

Find the real roots of the following equations : 

2. a r 3 _2x-5 = 0. 5. x 3 - 17 x 2 + 54 x -350 = 0. 

3. a* + x 3 - 500 = 0. q a; 4 -4 a 3 -3 a: + 27 = 0. 

4. x 3 - 7 a; +7 = 0. 7. x 4 - 12 x 1 + 12 x -3 = 0. 

APPROXIMATION BY DOUBLE POSITION. 

551. Find two numbers, a and b, the one greater and the 
other less than a root of the equation (Arts. 517 or 521), and 
suppose a to he nearer the root than b. Substitute them 
separately for x in the given equation, and let A and B repre- 
sent the values of the first member thus obtained. If a and b 
were the true roots, A and B would each be ; hence the 
latter may be considered as the errors which result from sub- 
stituting a and b for x. Although not strictly correct, yet, 
for the purpose of approximation, we may assume that 

A : B = x — a :x — b 

Whence (Art. 348), A — B : A = b — a : x — a 

or (Art. 345), A — B : b — a = A:x — a (1) 

A(b-a) 

and, x — a = — — 

A — B 

A(b-a) 
or, x = a + A _ B . 

From (1), we see that, approximately, 

As the difference of the errors is to the difference of the two 
assumed numbers, so is either error to the correction of its 
assumed number. 

Adding this correction when its assumed number is too 
small, or subtracting when too large, we obtain a nearer 
approximation to the true root. This result and another 



418 ALGEBRA. 

assumed number may now be used as new values of a and b, 
for obtaining a still nearer approximation ; and so on. 

It is best to employ two assumed quantities tbat sball differ 
from each other only by unity in the last figure on the right. 
It is also best to use the smaller error. 

This method of approximation has the advantage of being 
applicable to equations in any form. It may, therefore, be 
applied to radical and exponential equations, and others not 
reduced to the general form (Art. 480). 

EXAMPLES. 

1. Find a root of the equation x z + x- + x — 100 = 0. 

When 4 and 5 are substituted for x in the equation, the 
results are — 16 and + 55, respectively; hence a = 4, b = 5, 
A — — 16, and B = 55. According to the formula,' the first 
approximation gives 

_ 16 (5 - 4) , 16 , _ 
* = 4+ -16-55 - 4 + 7l = 42+ - 

As the true root is greater than 4.2, we now assume 4.2 
and 4.3 as a and b. Substituting these values for x in the 
given equation, we obtain — 4.072 and + 2.297 ; therefore 4.3 
is nearer the true root than 4.2. 

„ , Q 2.297(4.3-4.2) .2297 

Hence, x = 4.3- ^ + 4 _ = 4.3 - -^ 

-4.3 -.036 = 4.264. 

Substituting 4.264 and 4.265 for x, and stating the result 
in the form of a proportion, we have 

.0276 + .0365 : .001 = .0270 : correction of 4.264. 

Whence the correction = .00043+. 

Hence, x = 4.264 + .00043 = 4.26443+, Am. 

Find one root of each of the following equations : 



SOLUTION OF HIGHER NUMERICAL EQUATIONS. 419- 

2. x 3 - 2 x- 50 = 0. 4. a; 3 + 8r+ Gx — 75.9 = 0. 

3. * 3 +10x 2 +5a;-260 = 0. 5. x 3 +^-^- T = 0. 

lb 4 

6. x 4 - 3 x 2 - 75 x - 10000 = 0. 

7. a; 5 + 2 x 4 + 3 x 3 + 4 a; 2 + 5 x - 54321 = 0. 

NEWTON'S METHOD OF APPROXIMATION. 

552. Find two numbers, one greater and the other less 
than a root of the equation (Arts. 517 or 521). Let a he one 
of those numbers, the nearest to the root, if it can be ascer- 
tained. Substitute a + y for x in the given equation ; then y 

is small, and by omitting y 2 , y 3 , , a value of y is obtained, 

which, added to a, gives b, a closer approximation to the value 
of x. Now substitute b + z for x in the given equation, and a 
second approximation may be obtained by the same process as 
before. By proceeding in this way, the value of the root may 
be obtained to any required degree of accuracy. 

The assumed value of x should be nearer to one root than 
to any other, in order to secure accuracy in the approxima- 
tion. 

EXAMPLES. 

1. Find the real root of the equation x 3 — 2 x — 5 = 0. 

When 2 and 3 are substituted for x in the equation, the re- 
sults are — 1 and + 16 respectively ; hence a root lies between 
2 and 3, and near to 2. Substitute 2 + y for x, and there 
results 

y 3 + 6y" + 10y-l = 0. 
Whence, approximately, y = .1. 

Now substitute 2.1 + z for x, and there results 
.061 + 11.23 2 + = 0. 



420 ALGEBRA. 

Whence, approximately, z = ' i}0 = — .0054, and 

x = 2.1 - -0054 = 2.0946, nearly. 
Find one root of each of the following equations : 
2. a; 8 — 3 x -(- 1 = 0. 3. x s - 15 x- + 03 x - 50 = 0. 



ANSWERS TO EXAMPLES. 



In the following collection of the answers to the examples and problems given in the 
preceding portion of the text-book, those answers are omitted which, if given, would 
destroy the utility of the problem. 



Art. 47; pages 10 and 11. 

1. 93. 5. 408. 9. 5§. 13. 36. 17. llf. 

2. 136. 6. 254. 10. 13$. 14. 48. 18. 9. 

3. 127. 7. 24. 11. 4. 15. 3. 19. 10. 

4. 156. 8. 310. 12. If. 16. 4. 20. 76. 

Art. 60; page 18. 

6. Ua-9vip 2 . 7. x. 8. Sab-lcd. 10. 3mn 2 -2x 2 y. 
11. 39 a 2 -2i ab + 5 h\ 12. - « + 3 c + 2. 

13. x-y+3m + 3n. 14. 3a + 3b + 3c + 3d. 15. x. 
16. n+ r. 17. 6mn — ab — 4 c + 3x + 3m 2 — 4 p. 
18. 4 a - 2 b - 12 - 3 c - d + 4 x 2 - 18 m. 19. 6 a 3 . 
20. 14 six. 21. 7 a b + 7 (a + b). 22. 16 \/ y - 4 (a- b). 

Art. 66; page 21. 

6. — 3a5+4ctf— 5 a x. 7. 6 z + 12 ?/ — 8 « + 4. 

8. _4a6c-14a--2y- 148. 9. 2 y/ a - 4 y- + 12 a .+ 1. 
11. 14 x 2 -Sj/ 2 + 5ab-7. 12. 2 b -2 e. 13. 6 b + 1. 

14. 4m^8»-r+3s. 15. Qd -2 b — 3 a -3c 
16. 5m 2 +9w 3 -71a;. 17. 2 b. 18. a -6 -3 c. 



422 ■ ALGEBRA. 

Art. 74; page 24. 

4. a — b + c + cl — e. 5. 2 a + 2. 6. x — y. 7. a — 3b + c. 

8. 5wr-6»-4(i. 9. 6?» — 3%. 10. 4x + 2y. 

11. —35 — 7c. 12. a — c. 13. 9a+l. 14. 6 m + 2. 

Art. 86; pages 29 and 30. 

3. 6a 3 -16a 2 ?/+6ay 2 + 4?/ 3 . 4. a 4 + 4a + 3. 5. a 2 -b 2 
+ 2bc-c\ 6. -0a 2 +16a&-8£ 2 . 7. 6 8 -a 8 . 8. o 4 -«. 
9. 30 a 3 - 43 a 2 b + 39 a b 2 - 20 b 3 . 10. 6 a 4 + 13 x 3 - 70 a 3 
+ 71a-20. 11. -a 5 - 37a; 2 + 70 x -50. 12. -Gar 5 - 25 a 4 
+ 7a 3 + 81a 2 + 3a-28. 13. 2a 5 b 2 -3a 4 b 3 -7 a 3 b 4 + 4:a 2 b s . 
14. 4x 2m+1 y 3 — 1G x m+6 y n + 1 + 12 x 5 y 2n ~\ 15. 12 a; 6 +7 a; 4 
+ 5 x 3 + 10 x — 4. 16. m 5 + re 5 . 17. a 5 — 5 a 4 6 + 10 a 3 b 2 
-10 a 2 b 3 + 5 ab 4 - b\ 

Art 87; page 30. 

2. 6 a 2 + 11 a J + 4 6 2 . 3. a 5 + x 5 . 4. a 8 - 2 a 4 a 4 + x 8 . 
5. 2 a m + 1 - 2 a n + 1 - a m + n + a 2n . 6. 1 - a 8 . 7. a 3 + 3 a 2 a 
- 10 a x 2 - 24 x 3 . 8. a 5 -5 a 4 + 10 a 3 -10 a 2 + 5 a- 1. 

Art. 101; pages 37 and 38. 

3. a x - 2. 4. 3 b 2 - 4 a 2 . 5. 4 « 2 - 3 b 2 . 6. 3« 4 + 3a 3 i 

+ 3a 2 6 2 + 3«6 3 + 36 4 . 7. a 2 -ax + x 2 +-^—. 8. x 3 -x 2 y 

a + x 

+ xy 2 -y 3 + —¥—. 9. 2 x 2 - 7 a - 8. 10. 5 x 2 - 4 a + 3. 
a; + y 

11. x 2 - 2 x - 3. 12. a 4 + x 3 y + a- 2 ?/ 2 + x y 3 + y\ 

13. 3a 3 -2a 2 + a-5. 14. 2 a- 3 - a- + 1. 15. a-b+c. 

16. a 2 — 3 a— y. 17. x + //. 18. a n — b m + c r . 

19. l + 2a + 2a 2 + 2a 3 + ... 20. a-ax + ax 2 -ax 3 + ... 



ANSWERS TO EXAMPLES. 423 

21. a*-an + a i b 2 -ab* + b i . 22. 2 a 3 - 2 a 2 - 3 a - 2. 

23. - x 2 - 2 x - 4. 24. a 3 - a: + 2. 25. 2 a 2 - a b + 2 & 2 . 

Art. 107; page 40. 

23. 1 - a 2 + 2 a b - b 2 . 24. a 2 - b 2 - 2 b c - c 2 . 

25. a 2 -2ab + b 2 -c 2 . 26. c 2 - a 2 + 2 «6 -£ 2 . 

27. a 2 + 2«i + i' 2 -c 2 + 2^- c/ 2 . 28. a 2 - 2 a 5 + 5 2 

-c 2 +2crf-£ 29. a 2 + 2 a & + & 2 - c 2 - 2 c d - d 2 . 

Art. 115; page 42. 

3. ( a + x )(b + y). 5. (x + 2)(x- y). 7. (x 2 - y 2 ) (m - n). 

4. (a-m)(c + tf). 6. (a-b)(a 2 + b' 2 ). 8. (a: + 1) (x 2 + 1). 
9. (3 a; + 2) (2 a; 2 -3). 12. (ab-cd) {ac + b d). 

10. (2 a; - 3 y) (4 c + d). , 13. (m 2 x-ny) (n 2 x - m y). 

11. (2-7m 2 j(3»-4m). 14. (4ww-7a;y)(3a& + 5crf)- 

Art. 117; page 45. 

9. (a + b + c + d) (a + b — c — d). 
10. (a — c + b) (a — c — b). 11. (m + x—y) (m — x + y). 
12. (x — m + y — n) (x — m — y+ n). 

16. (x + y + 2)(x + y-2). 18. (3c + d+l)(3c + d-l). 

17. (a + b-c)(a-b + c). 19. (3 + x 2 -2y) (3-x 2 + 2y). 

20. (2 a — 5 + 3 d) (2 a - J - 3 d). 

21. (2 m 2 + 2 b - 1) (2 m 2 - 2 b + 1). 

22. (a — m + b -f- n) (a — vi — b — n). 

23. (a + m + b — n) (a + m — b + n). 

24. (x — c + y — d) (x — c — y + d). 

Art 118; page 49. 

25. (x 2 - 24) (x 2 - 5). 27. (x y 3 + 12) (x y 3 - 10). 

26. (c 3 + 11) (c 3 + 1). 28. (a b 2 - 16) (a b 2 + 9). 



424 • ALGEBRA. 

29. (x + 20 n) (x + 5 n). 32. (x + y - 5) (x + y — 2). 

30. (m 2 + 11 n 1 ) O' 2 - 6 ir). 33. (x - 8 y 2 g) (aj + 6 y 2 s). 

31. ( a _ 5 _ 4) ( a _ 5 + 1). 34. (m + n + 2) (m + ra - 1). 

Art. 121 ; page 53. 

3. Sab (a -' r 2) 2 . 7. 3 a 2 (a - 5) (a - 2). 

4. 5 x y 2 (3 x - 4 y 2 ) 2 . 8. 2 c m (c + 7) (c - 3). 

5. 2ay(3a; + y)(3a; — y). 9. a; y (m — 6) (m + 2). 

6. a; (x + 7) (a; + 1). 10. ±ab(2a + b) (±a 2 -2ab + b 2 ) 

11. (n - 1) (>r + » + 1) O 6 + w 3 + 1). 

12. (x 2 + y 2 )(x-t- 2 /)(^- 2 /). 

13. (cc 4 + m 4 ) (a; 2 + m 2 ) (x + m) (x — m). 

14. (m + w) (??i — w) (m 2 + m w + ?z 2 ) (??i 2 — mn + ri 2 ). 

15. (a + c) (a 2 - a c + e 2 ) (a 6 - a 3 c 3 + c 6 ). 

16. (2 a + 1) (2 « - 1) (4 a 2 + 2 a + 1) (4 a 2 - 2 a + 1). 

Art. 125 ; pages 55 and 56. 

Z. ax. 6. x + 7. 9. x (x — 1). 12. 2 a: + 5. 

4. m + n. 7. 2 a; — 3. 10. a — 2 &. 13. m(j;- 1). 

5. x 2 + 1. 8. 3 a - 4. 11. x + 6. 14. 4 a? - 1. 

Art. 126 ; page 61. 

6. 2 x + 3. 10. 2 a; - 5. 14. x 2 + x + l. 

7. 8a; — 7. 11. 5x + 3. 15. a — a;. 

8. x - 1. 12. jb + 2. 16. x 2 - 2. 

9. 3 a; + 4. 13. 2 a; - 1. 17. 2 (x + y). 

18. 2«-3x. 19. 3 a; + 2. 

Art. 130; page 63. 

2. 120a 4 6 2 c. 4. 36 a & b\ 6. 840 a 2 c 2 d 3 . 

3. 30xV« 8 . 5. 480 m 3 rt 2 x 2 t/ 2 . 7. 252 a 8 y 3 s 8 . 

8. 1080 a 2 b 2 c 3 d\ 9. 168 m n 2 x z y s . 



ANSWERS TO EXAMPLES. 425 

Art. 131; pages 63 and 64. 

2. ax(x-\-a)(x—a)(x-+ax+a 2 ). 7. ax (x — 3) (x— 7) (x + 8). 

3. 12 abc (a + b)(a-b). 8. (2x+l)(2x-l)%4:x 2 +2x+l). 

4. «(cc + l)(a;-l)(x 2 — rc + 1). 9. 3 ab (x — y) 2 (a — b). 

5. 24(l + cc)(l-x)(l + a: 2 ). 10. 2az; 2 (3;c + 2) 2 (9a; 2 -6x+4). 

6. (x+l)(a;-2)(cc+3)(*+4). 11. (x-l)(x-3) (x + 4)(x-5). 

12. (x + y + z) (x + y — z) (x — y + z). 

Art. 132; page 65. 

2. (3x-4)(4 : r-5)(2a; + 7). 4. (a 2 -2a-2)(a + 3)(2a-l). 

3. (4a;+l)(2a;+7)(3a;-8). 5. (2x + 3)(x 2 -x + l){x 2 +x-2). 

6. (a — b)(a 2 -ab + b 2 )(a 2 + 2ab + b 2 ). 

7. a x (x + 1) (x 2 — x — 1) (x 2 + x + 1). 

8. x (x - 5) (2 x 2 - x -2) (3 x 2 + x - 1). 

If the above expressions are expanded, the answers take the 
following forms : 

2. 24 z 3 + 22 a 2 -177 a; + 140. 4. 2« 4 + a 3 -17« 2 -4« + 6. 

3. 24 z 3 + 26 z 2 - 219 a; -56. 5. 2x 5 + 3x i -±x 3 + 5x-Q. 

6. a 5 -a 3 b 2 + a 2 b 3 -b\ 

7. a x G + ax 5 — ax 4 — 3 ax 3 — 3 ax 2 — ax. 

8. 6ic 6 -31a; c -4x 4 + 44a; 3 + 7a; 2 -10a;. 

Art. 148 ; page 71. 

14 - *£■■ 18 - 

o + 2 c 
a (2 + 3 n) 
1D * b{2-3n) • 19 ' 

16. ±*--J*y+v\ 20 . 

m + 9 3 y — 5 



10. 


cd 
3 xy' 


11. 


x a 
2?' 


12. 


sc — 5 


# + 7' 


13. 


m — 2 



2 


x 2 y 


5 


— X 

2 + x 


X 

c 


(J-x) 
-d 


c 

III 


+ d' 



. — nr 



m 2 — n 



426 ALGEBEA. 

Art. 149; page 72. 

2. 3 * 



3. 



4 a; + 1 

5a + 7 
a-2 ' 

10. 





4. 


m — 1 




6 m — 5 




5. 


x + 2 
x — 3 " 


x' 


2 


3a; + l 



x' z — x + 3 

Art. 150; page 73. 



6. 


3 a; — 2 


8 2a; + 5 
2 x -7 


a; + 3 * 


7. 


2x-3 


_ C « — 1 

"• 7= » 

5a — 7 


2a;-5* 




, 1 2x 2 -a;-2 




U< 2 a 2 


+ 3a; + l* 



3 


a - 


a* 




4. 


a; 2 


— X 


!/ + */• 


6. 


a; 2 
3" 


a; 
~~3 


7 
+ 3" 


2 
a;" 


7. 


a 
2b' 


3 2b 

2 + a ' 



K 2x 3 
o 



5 5 5a; 
23 



8. 2a; + 6 + 



9. x 2 + a; + l. 10. 2 + 77 -^- — ^-, 11. a;-24 

12. jc 



x — 3 
2a;-4 

2x z — x+1 ,x z +x — 1 

a; — 2 
2a; 2 -3a; + 3' 

Art. 151; page 74. 

_ (x — 1) 2 _ an + b-—cd 56.r — 4?i 2 — 5a 

2. =— . o. . 4. - . 

x — 6 n o 

(a; + l) 2 . 2ab „ a 2 +2b 2 Q a 3 + b 3 

0. . b. j. 7. — s . o. — . 

x a+b 2 a a — b 

Q 6a; 2 -7a;-l 2b 2 ,, a; 3 -2a; 2 -3a; 

"• ^ : n • I". — ■ -. 11. „ . 

2a; — 1 a+& x — 2 

Art. 152; pages 76 and 77. 

27 a b 16 a c 30 6 n 3 a; 2 ?/ 2 a-?/,? 7 y z 2 

3 ' 772~' "T^ 7 ~72~" 30 ' 30 ' 30 ' 

18 ?/a 2 16 x 2 z 2 15 a; 2 y 2 ' 
12 xyz* 12xij z' 12 xyz' 
40c 2 -10c 18i 2 -12 6 25 « a _2^_ 3_o^ 4 a a; 2 

30 a&c ' 30 abc ' 30 a be' ' a*x 8 ' a 8 a; 8 ' a 8 a; 3 ' 



8. 



ANSWERS TO EXAMPLES. 427 

100 a y z 3 Aob x 3 z 8Acxy s — 12mx y 2 
120 x 1 ?fz 2 ' 120 x 2 y 2 z 2> ' 120 x 2 y 2 z 2 * 



(a+b) {a 2 + U 1 ) (a - b) (a 2 + b 2 ) a 2 - b 2 



10. 



a 4 -6 4 at-b* ' at-W 

x 2 — 9 x 2 -l x 2 -4 



(x-l)(x-2)(x-3)' (x-l)(x-2)(x-3)' (x-l)(x-2)(x-3)' 



2 a (a + 2) 3 b (a — 2) 4 c (a + 3) 



12. 



0-2)(a+2)0+3)' 0-2)0+2)0+3)' (a-2)(a+2)(a+3)' 
x 3 +2x 2 -\-2x + l x 2 + x + l x + 1 



(x + 1) {x 3 - 1) '0 + 1) (X s - 1)' {x + 1) (a 3 - 1)' 
13 6 a 2 b 2 3 b (m 2 - n 2 ) 2 a (a 2 - b 2 ) 



6ab(a — b)(m + ?iy 6ab(a—b)(m + n)' 6ab(a—b)(m + ii)' 

3 Q+l) 2Q-1) 2-a 1-x x 2 -x-2 3 

' a»-l' a 2 -l'a 2 -l' "*' l-x» 1-jb 2 >1=?' 

17 c2 -^ 2 Q-l)Q + &) Q-6) 2 

' O 2 - 6 2 ) (c - d)' O 2 -* 2 ) (e-d)' (a 2 -b 2 ) - d)' 

Art. 153; page 78. 

2 ( a ~ b y 3 g 2 +9«+8 9 m 2 - 4 

a* — b 2 ' ' x 2 +5x — 24' "6m 2 — 19m + 10" 

4 2 +«& + 6 2 ) 1_^ 

a 3 — & 3 1 — x 

Art. 154; pages 80 to 82. 

4 12 a; + 7 6a + 5& a + 3 3 m 2 n 2 - 4 

36 ' ' 10 a 2 6 2 ' 24 • 6m 2 -» 3 * 

8 5 &* + 4 «' q 5a + 6 in 4,ab-b-±a 3 1 

" 120 a 6 " * 24 • iUl 12« 3 T~ ~* 1L 15" 

12 — is 3a; ~2 1 45ctf+6acd— 3ahd— 2abc 

•42' "■ 18 x 2 ' 14l_ 60- 15 -~ ~48^X~ 

17. ? 18 1 19 2 Q^ + & 2 ) 

6 + x-x 2 ' ' x 2 +15x+56' a 2 -b 2 ' 



428 ALGEBKA. 

■Oi *" «l£±|. 22.. 4^. 23. (* + 2 y 



1_^' — a-6* -a 4 -^' • (.x + l)(a; 3 -l) 

13-18* _J_ a 2 -14a + l 

^ (z + lXa^Xz-S)* 0, &-«' ' 6(^-1) 

28. 57 _?-_. 29. 44- 30. 0. 31. 



9 a:- a; 3 ' ' »*— 1" ' ' ' (»-2)(jb-3)(»-4) 
Art. 155; pages 83 to 85. 



2. 


a 5 b s c 
m* n 3 d' 


3. 


12 a 4 b x 
35 h 5 m ' 




4 ^ 


7. 


/ • 8 - 

4 a.' y 


a 3 . 


9 11 


o 

m 
4 


w 2 3 a; 

a; - 


-1 10* 
-2 * "■ 3 


13. 


b(a-b) 
x(a + b) 


14. 


a — b 
a 2 ' 




I*' 1 -'. 

a; 


ia x*-x-2Q 
16 ' 


17. 


a (x-2) 
a + 1 


18. 


X 

x-2' 




19. "+*• 

. or 


20. aj2 + 5 f + 6 

ST 



21. -^o . 22. x- + xy. 23. 1. 24. 2. 25. -J£-j. 

ic + 2 % ~r y 



. 91 ra 2 
o. 



6rc 2 

2 



4. 



Art. 156 


; page 86. 




5 mx 


7.!. 

X 


9.» + 1 ' 

a + 5 


, 3 0-4) 


8 3.t- 2?/ 


10. aj. 



8 b 3 m ri 1 x s x + y 



Art. 157; pages 88 and 89. 



a 



_ acn + bn „ 3m— n „ Ay—±x + 2a 

— — . o. — — . b. — x . «. ^j • 

bm + b?i cnx — cm ox oi 

x 2 y 3 +l 



8. x — 1. 9. a; 2 -rK + l. 10. a + b. 11. 



xif — ly* 



, n a— J 10 «— 4 ... 1( . 4 R aft 

12. -T7-. 13. ^. 14. aj. 15. k— -Q- lb - ■> , 7,2 ■ 

a- b x + 6 3 a; + 3 a" + 



ANSWERS TO EXAMPLES. 429 



17 1 18 ww(m-w) 8 19 _ s-g 

m* + m 1 ri 1 + w 4 ' x + 2 a 



Art. 175; pages 94 and 95. 

2. aenx—bcen=bdnx—bem. 3. 6 bx — 8a 2 = 3 — 2a6a. 

4. bdex— adex+bcex— abd=0. 5. \2x-\-hx — 6x— 1320- 

6. 9z-12a=10a:+24-46. 7. 28a:-4a;+560=14a;+7a;+728. 

8. ±ax-6c-5a s x + 2a 3 bd=0. 9. 10x-32a;-312=21-52a;. 
11. 3 a — 2 a — 2 a; = 45. 12. a 6 a; + b 1 — ex — d = a c. 

13, 3-3z-2-2x = 0. 14. 6ar+3z-6ar+ 18-4a;-2=0. 
15. 3 a ._3_2a--2-5x=0. 16. 6x + 6-15aj + 45-20a;-10=0. 







Art. 


177; pages 


97 to 102. 




4. 


3. 


13. 1. 


22. 72. 


31. 5. 


41. - 1&. 


5. 


7. 


14. 2. 


23. 60. 


32. -5. 


42. 4£. 


6. 


-1. 


15. 2. 


24. 10. 


33. 4. 


43. l t V 


7. 




16. -4. 


25. -2* 


34. -5. 


44. 0. 


8. 


1. 


17. 2. 


26. 56. 


35. -2. 


45. H. 


9. 


3 

< 


18. 1. > 


27. ?. 


33. f . 


46.-^. 


11. 


3 

— 2' 


20. 3. 


29. -2. 


37. -1. 


47. 1. 


12. 


0. 


21. 5. 


ao.J. 


40. -7. 


50. f -* 

2a+b 


51. 


a 2 

a 2 - 


+ 4a . 52 
3a+2 


!. 2&. 53. 


If *I 


Abc + a* 
a?—b + 16c 


55. 


2 a 2 
36 


, 57. ^. 




59. \ 

a + 2 


i- 61 --^ 


64. -3. 


56. 


a. 

r 


58. 12 a 3 . 


60. aft. 


63. 2. 


65. 50. 






66 B5- 


67. 5. 


68. 0. 





430 ALGEBEA. 

Art. 182 ; pages 108 to 113. 
10. Horse, $224; chaise, $112. 11. 37. 12. 10 and 7. 

13. 18 and 2. 14. 58J and 4l£. 15. A, 40 ; B, 20. 

16. A, 60 ; B, 15. 17. If. 18. ft. 19. 23£. 

20. 84. 21. 36. 22. Oxen, 12 ; cows, 24. 

23. Wife, $864; daughter, $288; son, $144. 
24. Worked, 20 ; absent, 16. 25. Horse, $ 126 ; saddle, $ 12. 

26. Infantry, 2450 ; cavalry, 196 ; artillery, 98. 
27. 144 sq. yds. 28. Water, 1540 ; foot, 880 ; horse, 616. 

29. $ 1728. 30. $ 2000 at 6 p.c. ; $ 1200 at 5 p.c. 31. 7. 
32. 31. 33. $24. 34. $100. 35. 142857. 

36. A, $ 466| ; B, $ 533J. 37. 2 dollars, 20 dimes, 4 cents. 
38. $2.75. 39. Men, $25; women, $21. 40. 23 and 18. 
41. 48 minutes. 42. 12121 men ; 110 on a side at first. 

43. 5 r \ minutes after 7. 44. 43 T 7 T minutes after 2. 

45. 27f\ minutes after 5. 46. 29 and 14. 

47. 3377 ounces of gold ; 783 ounces of silver. 48. $ 2000. 

49. 30 bushels at 9 shillings ; 10 at 13 shillings. 50. 10 a.m. 
51. $ 1280. 52. 21 ft minutes, or 54 T <Y minutes after 7. 

53. 27^ T minutes after 4. 54. 23*1 miles. 

55. Greyhound, 72 ; fox, 108. 56. 1 minute, If §f seconds. 

Art. 192 ; pages 120 to 123. 

3. x = 4, y = 3. 9. x=-2, y=10. 15. x = 12, y = 18. 

4. s = 5, y = -2. 10. jk==12, y = S. 16. a; = 35, y = -10. 

5. x = 7, y = 5. 11. x=~2, y=-10. 17. x = - 28, y = 21. 

6. x = - 8, y = 2. 12. x = 10, y = 5. 18. x = A,y = .1. 

7. x = 5, y = 7. 13. x = 7, y = 11. 19. x = l£, y = 3f . 

8. x=-8, y=-12. 14. a>=ll, y=-9. 20. x = 3, y = -2. 
21 dm — bn _an — c m n / + ri r 

ad — be ad — be ~ mri + m' n ' 



ANSWERS TO EXAMPLES. 431 

m'r—mr' nn ac(bvi + dv) bd(cn-am) 

V = • "3. X = ^-= ; -, y = ^ : -. 

u m n' + w! n ad + b c ad + b c 

11 25 5* 

24. x=—, y=-^. 25. x = 60, y = 40. 26. x = — , y=». 
2 a 2, a Ob 

27. x = lft, y = 4 T V 28. x = -6, y=-5. 30. # = 4, y=2, 

be — ad 

bn — d vi ' 



31. x = — 5, y=3. 32. x = -2, y — — l. 33. ■ x = 



b c — a d „. 3 2 oe 1 1 

y = • 34.33 = ^77,2/ = —^. 35. x=-,y = — 

cm — an a~ b a tr n vi 



Art. 194; pages 126 and 127. 

3. x = 23, y = 6, g = 24. 6. a; = — 5, y = — 5, z = — 5. 

4. x. = — 2,y = 3, z = l. 7. « = 4, x = 5, y = 6, z = 7. 

5. x = 8, y = - 3, g = - 4. 8. a; = 3, y = - 1, s = 0. 

9. jc =^ (b + g — a), y = ^ (a + c — b), y = ^ (a + & — c). 

10. x~^,y = 7 -£,z=^ 11. a = -24,y=-48,g=60. 

12. u = — 7, x = 3, ?/ = — 5, z = 1. 

_ 5 2 +c 2 -ft 2 _a 2 +c 2 -& 2 _ a 2 + & 2 -c 2 

13, *~ 2bc ~ ,V ~ 2ae ,Z ~~ 2ab ' 

14. x = -,y = -^,z=--. 15. a; = li y = -X\, «='l. 

16. x = ab c, y=ab + ac + bc, z = a + b + c. 

a + 1 f- 

17. x = 7, y — — 3, g = — 5. 18. as = ,y—a — c,z-- 



c a 

Art. 195 ; pages 129 to 133. 

4. A, 30 ; B, 20. 5. ^. 6. Cows, 49 ; oxen, 40. 

7. A, $140; B, $70. 8. A, 98; B, 15. 9. 32 and 18. 

10. Man, 24 ; wife, 18. 11. Worked, 6 ; absent, 4. 

12. Horse, $96; chaise, $112. 13. A, $96; B, $48. 



432 ALGEBRA. 

14. 16 days. 15. 13J bushels at 60 cts. ; 26| at 90 cts. 
16. Wheat, 9; rye, 15. 17. Income tax, $20; assessed tax, $30. 
18. A, $500; B, $700. 19. 30 cents ; 15 oranges. 
20. 1st, 8 cts. ; 2d, 7 cts. ; 3d, 4 cts. 21. Better horse, $40 ; 
poorer, $30; harness, $50. 22. 10, 22, and 26. 23. 246. 

24. A, $2000; B, $3000; C, $4000; D, $5000. 
25. A, 45; B, 55. 26. A, $20; B, $30; C, $40. 

27. Whole sum, $120; eldest, $40; 2d, $30; 3d, $24; 4th, $26. 
28. Length, 30 rods ; width, 20 rods ; area, 600 sq. rods. 
29. Going, 4 hours ; returning, 6 hours. 
30. A, 9f days ; B, 16 ; C, 48. 31. 1st rate, 6 p.c. ; 2d, 5 p.c. 
32. 15 miles ; 5^ miles an hour. 33. 30 miles an hour. 

34. A, 5 ; B, 6. 35. First, 22 ; second, 10. 36. A, 8 ; B, 6. 

Art. 197; pages 136 and 137. 

. a b c _ . , _ m a .. n a 

4. — — . 5. li hours. 6. and . 

ab + ac + bc vi-\-n m,-\-n 

7. 12 and 8. 8. -^-. 9. 12. 10. 10 °" 11. $2100. 

b — a rt+ 100 

12. 100 >-i>), 13. m. 14. lst, g(c - 6) ;2d ; ft(a - r) . 

p r ~ a — b a — b 

,.-.». ^,.,/n , „ b + d ^„ am+bji + cp 

15. 1st kind, 5 ; 2d, 10. 16. — — . 17. - -r- — . 

a — c a + b+c 

* amt . -o ant ' n a P t " 



mt + nt' -\-pt' n ' mt + nt'+ptf n ' mt+nt'+pf' 
Art. 205; page 141. 



5 



3.-2 rods. 4. - ^ . 5. 105 and — 15. 6. In — 30 years. 

7. A, .-$1500; B, -$500; that is, A was in debt $1500, 
and B $ 500. 8. Man, $ 3 ; son, — $ 0.50 ; that is, the man 
was at an expense of 50 cents a day for his son's subsistence. 



ANSWERS TO EXAMPLES. 433 

Art. 225; page 152. 

4. x > 5. 5. x > 15, x < 20. 6. 4. 7. x > 6|, y > 2|. 

8. a; > c, x < d. 9. * > 9|, 7/ < 12£. 10. 19 or 20. 

11. Any no,, integral or fractional, between 8 and 15. 12. 60. 



Art. 229; page 155. 

2 7 2 

1. a 3 -3a 2 b + 3ab 2 -b 3 . 2.~-2+— i . 

3. 1 + 3 a 2 + 3 b 2 + 3 a 4 + 6 a 2 6 2 + 3 b* + a & + 3 a 4 b 2 + 3 a 2 6 4 + 6 6 . 

4. a 2 + 2am — 2an + m 2 — 2mn + n 2 . 

5. a *m _ 4 a 3m + », _|_ (5 a 2m + 2n _ 4 (( m + 3a + a 4 n# 

6. a 5 + 5 a 4 5 + 10 a 3 b 2 + 10 a 2 b s + 5 a 6 4 + 6 5 . 

Art. 230 ; page 156. 

3. 4a: 4 +12ar+25ar 2 +24z + 16. 4. 4a: 4 -12a; 3 + lla; 2 -3a: + |:. 
6. a; 6 +4a: 5 +6a; 4 +8a; 3 +9a: 2 +4a;+4. 7. l-4a;+10a: 2 -12a; 3 +9a: 4 . 

8. 1 + 2 a; + 3 ar + 4 x 3 + 3 a; 4 + 2 a: 5 + a: 6 . 

9. x 6 -8x 5 + 12 x* + 10 a; 3 + 28 x 2 + 12 x + 9. 

10. 4 x« + 4 a: 5 + 29 x" + 10 a; 3 + 47 a: 2 - 14 x + 1. 

11. a; 6 + 10 x 5 + 23 x i - 6 a: 3 + 21 x 2 - 4 a; + 4. 

12. 9 a; 6 - 12 x 5 - 2 x i + 28 x 3 - 15 a; 2 - 8 a: + 16. 

Art. 231; page 157. 

2. a^+Qa'b + 12 a 2 b 2 +U 3 . 3. 8m 3 +60?7r?z+150mw 2 +125» 3 . 

4. 27 a; 3 -108 a; 2 +144 a; -64. 5. 8 a; 9 - 36 a; 6 + 54 x 3 - 27. 

6. 64 x 6 -48 x 5 y +12 a; 4 ?/ 2 — a; 3 ?/ 3 . 

7. 27 a; 3 ?/ 3 + 135 a b 2 x 2 y 2 + 225 a 2 b A xy + 125 a 3 b 6 . 

Art. 232; page 158. 

3. a; 6 - 3 x 5 + 5 a; 3 - 3 x - 1. 5. 8 - 24 a; + 36 x 2 - 32 a: 3 
+ 18 x i - 6 a; 5 + a; 6 . 6. 1 + 3 x + 6 a; 2 + 10 a; 3 + 12 a; 4 



434 



ALGEBRA. 



+ 12 x 5 + 10 x« + 6 x 1 + 3 x 8 + x\ 7. 8 x» - 12 x s + 30 a; 7 

- 61 x* + GG x 5 - 93 a 4 + 98 a; 3 - 63 a 2 + 54 a; - 27. 



Art. 239 ; pages 162 and 163. 

2. 2 a; 2 - x - 1. 5. 3 - 2 a; + a; 2 . 8. 3 ar - 4 x - 5. 

3. 2 a 2 - 4 a + 2. 6. 5 + 3 x + x\ 9. 2 x 2 - 5 a; + 8. 
1 



4. m + 1 



m 



7. 1 — 7» -2 a; 2 . 10. a -b 



c. 



U. x-2y + 3z. 

13. a-+ S fr—, : 4 



a; 



a; 2 a; 3 



i 3 



8 a 3 16 a 5 
15. a + s 



12 - 1 + 2-^ + 16 






2 a 8 a 3 



+ 



x° 



16 a 5 









Art. 


241; page 166. 






2. 


523. 


7. 


95' 


12. 900.8. 


17. 


13.15295 


3. 


214. 


8. 


1.082. 


13. .4125. 


18. 


.88192. 


4. 


327. 


9. 


21.12. 


14. 1.41421. 


19. 


.43301. 


5. 


5.76. 


10. 


.083. 


15. 2.23607. 


20. 


.57735. 


6 


.97. 


11. 


.00328. 


16. 5.56776. 


21. 


.53452. 



1. 3.3166. 

2. 1.732051. 



Art. 242; page 168. 

3. 7.81024968. 5. 27.94638. 

4. 11.446. 6. 113.7234. 



Art. 243; pages 170 and 171. 
2.1-2?/. 4. 4:x-3ab. 6. y 2 — y-1. 8. a; 2 -2a; + l. 



3. 2ar + 3. 5. a: 2 + 2a;-4. 7. x + -. 

x 



9. a + b + c. 



10. 2 a; 2 



3a;-l. 



115 

11. X+'„ -o-K— R + 



3 a; 2 9 a; 6 ' 81a; 8 



12. 



x 





ANSWERS TO EXAMPLES. 


435 


a? 


a G 5 a? g 2 1 1 


5 


3x 2 


~9x 5 Six 8 '" . 4z 4 32z 10 


768x 16 "* 



Art. 245; page 173. 

2. 123. 5. «. 8. 1.442. 11. .855. 

3. .898. 6. 3.72. 9. 1.913. 12. .420.* 

4. 11.4. 7. .0803. 10. 5.963. 13. .561. 

Art. 247; pages 175 and 176. 
2. to 2 — 2m — 4. 3. a 2 -ax + x 2 . 4. 2x — 1. 5. x' 2 — x + 1. 

Art. 248 ; page 176. 
1. 2x-3y. 2. cr-1. 3. to 2 -2to-3. 

Art. 257 ; pages 180 and 181. 

4. c^. 5. af*. 6. *»*. 9. - 6 a J*K 

11. a 4 6~ 4 - 2 + a- 4 i 4 . 12. a - b. 

13. a- 5 -3a~ 3 b 2 + a- 2 b s -2a- 1 b i . 14. 18a 2 i 2 +10 + 2«- 2 &- 2 . 

15. 2 a r 1 2/-10z Z /- 1 + 8.ry- 3 - 16. 2-4x"V + 2affy. 
17. 6x 2 -7^-19af*+5a;+9^-2^. 18. 32aZr 2 -50+18 a - 1 & 2 



Art. 258; pages 182 and 183. 

5. <T*. 6. to 5 . 7. x 12 \ 8. ^ ' 

11. J + Jb^ + Jb^ + Jb^ + bl 12. a-^^^ + i- 2 
13. x- 3 f -x~ 2 y + 2x-\ 14. a>* y- 1 - 3 + 4 x~* y, 

15. x- 1 y- 2 -a;- 2 7/- 3 -x- 3 y- 4 . 16. 2x^y~ ?J -x~^y-x~^ >*. 

Art. 260 ; page 184. 
6. x*. 7. c~K 8. to- 1 . 9. 2T 3 . 10. a~ 3 . 11. w" 1 



436 ALGEBRA. 



Art. 262; page 186. 

' 1 n 12500 

3 - 9 - 5 -ioooo- 7 - 4 ' 9± -^~ 

4. ±216. 6. ±-^. 8.-243. 10. ±£. 



Art. 263 ; pages 186 and 187. 
5. 3 3-2 y _ 2 x- 1 - 2T 1 - 6. 2 aj^ + x y~^ -4^ y~*. 



./•' 



7. a?t"y T *-2-+aT*y*. 1L 2y*-y*ar 1 . 12. 
13. a;- 2 " 6 . 14. a*». 15. a 31 . 16. a - ^. 17. sc. 

01 •* + «* oo h^d'-afd* 9q 5 »" (»»-!) 

21 1_3« 3 * ^ ab*# + aP*' 3a ' 



Art. 267; page 189. 
2. ^27, #16, #25. 3. $625, ^216, ^49. 

12.^ 12, 12 „ 15,— — 15,-———- 15,. 



4. V*V, V^ 4 ^ VVs 3 - 5. #32 a 5 , #27 & 3 , #64c 3 . 



12. 12, 



6. #cr+2a6+6 2 , \/a 3 -3a-b+3ab 2 -b 3 . 7. #a°-3a 4 .z 2 +3a 2 ;c 4 -x 6 , 
fa 8 -2« 3 x 3 + x 6 . 8. y^3. 9. #2. 10. #4. 



Art. 269 ; page 190. 



5. \l$ab\ 6. #a& 2 . 7. v/(|f 3 )- 



Art. 270 ; page 191. 



11. 3xyS/2xy' 2 -3xhj. 12. (a — 3)^a. 13. (x + y)^x — y. 
14. (2 x + 3 a) f 6a. 15. 4 a & #3a& 2 + 5&. 18. ^ 6 - 
1 ._ „„ 1 ,„. „. 2a 



19. ^i/30. 20. Ji/21. 21. ^f y/3. 22. r> #6*. 

6 b y J 



ANSWERS TO EXAMPLES. 437 



a\j ab c 



2 (a + x) 6- (a + 6) 

Art. 272; page 192. 

7. v^x s- ft^F- 9- y/(^)- 

Art. 273; pages 193 and 194. 



«/. o 38 



3. 10 si 2. 4. 12 v/ 3. 5. 9^2. 6. ^y/o 

7- |\/6. 8. |j2 + ^18. 9.4^5. 10. i^- 

11.6asJ3a. 12. ^V 3 - 13. ^2 + ^3. 



14. 2 yV 2 - tf. 15. (2 a - 5 b) \Jl x. 

Art. 274; pages 195 and 196. 
5. \f^¥x-\ 6.^4500^ 7 - ^(g^ga)- 

8. y/5^ . 10. a; + v/z - 6. 11. 21 x - 38 v^ + 5. 

12.2. 13.-1. U. x-ij-z + 2\jy^. 

15.4 + 2^10. 16. 56+12v/35. 17.36-32^15. 

18. ax-x\ 19. m + TO. 20. 14 — 4^6. 

21. 147 + 30 v/24. 22. 1 + 2 a %/l-a 2 . 2-3. 2 a- 2 \/a 2 -& 2 . 

Art. 275 ; page 196. 

b/16 
V / 243' 



e/8 

V9- 



7. {/" 8. {18. 9. ^ 



438 ALGEBRA. 

Art. 276; page 197. 
3. ^125. 4. y/7. 5. 2304 x\ 6. a 4 * 2 . 7. \j~a~^-~b. 



8. SlaHx Sjbx. 9. a: 2 + 2x + l. 10. 16 a: 2 - 48. 



Art. 277; pages 198 and 199. 

12,- 



3.^2. 4. si 2. 5.<)a + b. 6. V* - 1. 7. y/2 

8. ^3. 9. v/3. 10. ^^V- n - V 2 - 



Art. 278 ; page 200. 

. 3s/2 . ^4V 2 x 5 \/ 2 a 2°\/3tf 

O. pr — . 4. — s . 0. — ^ — . D. 5 

2 2a 2 3a 



Art. 279; page 201. 

3. 12 ~ 4y/2 . 4. 5 + 2 V 3. 5. 2 ^ 6 - 5. 

a + 2_^b'+b 16 + 7 y/ 10 

6 ' " «-fi ^ ~* 7 * " 13 



_ a — 2 ^ ax + x a + 3 + 3 \la + 1 

a — x a 



a+^a*-x 2 



10. 2a 2 -l-2aV / « 2 -l- H. • 12. y^-l-ff 2 . 

a; 



1Q x*-2 + x^x 3 -4: 14 a: - 24- 11 Vs 8 - 2a; 

2 18 — o x 



Art. 281; page 202. 
2. .894. 3. 7.243. 4. 3.365. 5. .101. 

Art. 286; page 204. 



4. — 8 v/6. 5. 12\/ab. 6. 46. 7. 2. 

8. -abc V^l. 9. a 2 + b. 10. 12. 12. sJB. 



ANSWERS TO EXAMPLES. 439 

13. y/2. . 14. y/5. 15. ^3. 16. \f^l. 17. 1 + <f^2. 

18. 2 (f- b ) . 19. 1 _ 4 y/^3. 20. - 100 - 18 tf=2. 

or + b T 

Art. 293; pages 207 and 208. 



5. v/T + y/5. 8. 5 + y/lO. 11.^15-^5.14.3-2^-2. 



6. v/21-v/3. 9. 3-v/3. 12.3 + ^5. 15. 5 + 3 \/^2. 



7. 3 + v/7. 10. v^5-v/3. 13. 7-3 y/ 2. 16. 6-V-l. 



17. ^m + n— \jm — n. 18. x — SJax. 19. 3 + v/2. 

20. v/2-1. 21. 2-^3. 

Art. 297; pages 209 and 210. 



4. 17. 


9. 4. 14. 4. 19. -1. 24. 4. 


5. 19. 


10. 5. 15. 81. 20. -3. 25. 5. 


6. 7f. 


11. -2. 16. 4. 21. 4. 26. 3. 


7. 2. 


12. |. 17. 8. 22. 12. 27. 6. 


8. 4. 


13. 4. 18. -3. 23. 25. 28. 39. 


29. 


3£. 30. 3. 31. 6. 32. 3«-l 




Art. 303 ; pages 212 and 213. 


2. ±3. 


4. ±y/ (_?). 6. ±7. 8. ±1. 10. ± 


3. ±5. 


5. ±1. 7. iy/11. 9. ±1. 11. ± 




Ifr — IA 




12. ±^\L-±y 13. ±v/a + ft. 



Art. 310 ; pages 220 to 222. 
10. 5 or - 7. 11. 11 or - 2. 12. 5 or 3. 



440 ALGEBRA. 

13. - 5 or - 13. 29. - 4 or - 1. 45. 2. 

14. ^or-|. 30. 2 or i. 46. 4 or 0. 

15. 2 or -. ' 31. 4 or - If. 47. 3 or - 2. 

16,_l r-|. 32. 4±2y/3. 48. -2ov~ 
do bo 

17. ^^^ 33.3or-l. 49. ± 2 



12 --■ — ~ .y/8' 

18. 17± y 337 . 34. 2 or - 1 . 50. 25 or 3. 

4 7 

19. -- or -1 35. 7 or | . 51. 6 or -2. 

o 2 6 

20. lor- 7 , 36. 4or-^. 52. -or--. 

4 4 a. c 

21. =f or - 2. 37. - 10 ± v^78. 53. a ± b. 

o 

0Q 1±V409 3 a a 

**. £ . 38. — 3i or — 2^. 54. — — - or - . 

b ""42 

23. — or - . 39. 1 or — . 55. — a or — b. 

4 Z oh 

24. 3£ or - 1. 40. 1 or £ . 56. 11 or 18£. 

25. 13 or - 2. 41. 5 or ^ . 57. 5 or - 3. 

5 



26. I or i . 42. 18 or 3. 58. 12± J . 
2 14 5 

27. 1 or 3i. 43. — 2 or — . 59. a — ft or — a — c. 

28. - 4 or - ** . 44. - 3 or 2*. 60. ^=* or- 3 ^. 

_, a + b a — b 

bl. or — — r . 

a — o a + 



ANSWERS TO EXAMPLES. 441 

Art. 311 ; pages 224 to 227. 

4. 12 rds. 5. 40000 sq. rds., and 14400 sq. rds. 6. 9 and 6. 

7. 16 and 10. 8. 16. 9. 3 inches. 10. $ 30. 11. 14 and 5. 
12. $2000. 13. 18bbls., at $4 each. 14. 256 sq. yds. 15.5. 
16. 7 and 8. 17. 7, 8, and 9. 18. Length, 125 ; breadth, 50. 

19. 9. 20. 3712. 21. 80. 22. 20. 

23. Area of court, 529 square yards ; width of walk, 4 yards. 

24. 36 bu. at $1.40. 25. Larger, $77.17^; smaller, $56.70. 
26. 1st, 14400; 2d, 625 ; or, 1st, 8464 ; 2d, 6561. 27. 84. 
28. 6. 29. Larger pipe, 5 hours ; smaller, 7 hours. 

30. 38 or 266 miles. 31. 70 miles. 

Art. 314 ; pages 230 to 232. 

5. ±3or±V-13. 6. ±l or ±J_ 7. lor -2. 

I y 5 

8. ± 1 or ± - . 9. ± 7 or ± 5. 10. ^3 or - ^23. 

11. ± 8 or ± d(- ^ . 12. 4 or ^49. 13. 4 or 1. 

14. 243 or - J' (28 5 ). 15. 4 or 1\. 16. 49 or 25. 

18. 2, - 2, 3, or 7. 19. 3 or - 1. 20. ± 1 or ± 2. 

21. 2 or - 3. 23. 1, - 1, 5, or 7. 24. 2, - 3, 4, or - 5. 

25. 1, 2, - 5, or 8. 26. 1, - 1, - 6, or - 8. 



28. 3,-|or 3± 4 V/ 55 . 29. 8, -2, or 3 ± y/ 110. 

80.?,-?, or " 3± 2 2 ^ 3 . 31. 1,9, or 5±2 N /2. 

32. 0,-5, 1, or-^. 



Art. 317; page 234. 
2. x = 2 } y—±l; or, x = — 2, y=±l. 3. x = 4, y = ± 5 ; 



442 ALGEBRA. 

„ . 1 1 11 

or, x= — 4, y=±5. 4. « = g, y — ±^ ° r ' a;=— 3' ^ =± 2' 

1 1 

5. x = 3,y = ±p or,x = — 3,y = ±g. 



Art. 318; page 235. 

2. a? = 7, y = — 8; or, x = — 8, y = 7. 

3. a; = 5, y = — 2 ; or, a; = — 2, y = 5. 

4. x = 3, y = 4 ; or, cc = — 4, y = — 3. 

5 5 

5. x = S, y = 2> or > cc=: ~2' y:= — 8 * 

1 5 

6. x = 2, y = 4 ; or,x = — ^,y = ^. 

7. ar = 2, y = — 3 ; or, x = 3, y = — 2. 

8. x = 1, y = 2 ; or, a; = 2, y = 1. 

Q Q 9 15 62 

9. a? = 3, y — 2; or,x = — —,y = —. 

10. a? = 9, y = 6 ; or, x = — 6, y = — 9. 

11. a; = 2, y = 9 ; or, x = 9, y = 2. 

12. a; = 9, y = 3 ; or, x = — 3, y = — 9. 

13. x = 6, y = — 4 ; or, x = — 4, ?/ = 6. 

14. a; = 3, y = 2; or, a; = —, y = — — . 

15. x = 5, y = 3 ; or, a; = — 3, y = — 5. 

16. x = 3, y = — 7 ; or, a; = — 7, y = 3. 

Art. 319; page 238. 

4. x = 3, y = 4; a; = 4, y = 3; x = — 3, y = — 4; or, x = — 4, 
y = — 3. 

U. & = 6, y = 7 ; a; = 7, y = 6 ; a; = — 6, y = — 7; or, a; = — 7, 
y = — 6. 

6. x = 2, y = — 3 ; or, x = — 3, y = 2. 



ANSWERS TO EXAMPLES. 443 

7. x = — 1, y = 4 ; or, as = — 4, y == 1. 

8. # =1 3, y = — 2 ; or, x = — 2, y = 3. 

9. a; = 4, y = — 7 ; or, a; = 7, y = — 4. 

10. x = 5, y = 6 ; or, cc — 6, y = 5. 

11. x = 5, y = 2 ; or, a; = — 2, y = — 5. 

Art. 320 ; pages 239 and 240. 



2. x = 2, y = 2> x = - 2 ,!/ = -2> x = \ 5>y = — 2 \ 



2 
5 ; 



5 g 

3. x = 2,y = 3; x = — 2,y = — S- x = ^^ r ,y = 



^31 ' y ~ v' 31 ' 
5 6 

y/31' y "~^31.' 

4. aj = 3, y = l; a>= — 3, y = — 1; x = 2^2, y = \/2; 

or, ^ = -2^2, y = — ^2. 

5. £ = 3, y = 5;a; = — 3, y = — 5; x = -, y = y5 or, z = — -, 

13 

6. a; = 2, y = — l\ x = — 2,y = l-, x = —-r , y 



5 7 

7. a; = 2, y — 1 ; a; = — 2, ?/ = — 1 ; x = 7, y = — 19 ; or, 

x — — 7, y = 19. 

Art 321; pages 243 and 244. 

5. x = l, y = 8; or, x = 8, y = 1. 

6. a; = 4, y = 9 ; or, x = 9, y = 4. 

7. a: = 2, ?/ = 3 ; or, a: = 3, y = 2. 



444 ALGEBRA. 

8. x=3, y=4 ; 3=4, y=3 ; x=-± + \J^11, y =-A-\^H ■ 

or, x = — 4 — \/— 11, y = — 4 + y^— 11. 

9. a:=4,y=5;a:=16, y=-7 ; 3=- 12+^/58, y=-l— ^58 ; 

or, x =- 12 - y/58, y = - 1 + y/58. 

10. a; = 4, y = 2 ; cc — — 2, y = — 4 ; or, a; = 0, y = 0. 

11 o a 605 20 io -, 3 

11. x = 9, y = 4; or, *=.— , y = — . 12. a- = 1, y = -. 

13. x = 3, y = 2 ; or, a; = 2, y = 3. 14. x — 9, y = 4. 

15. 3 = 1, y=-3; a=-3, y=l; a-=l+V^^2, y=l- V^2; 

or, a; = 1 — \/-2, y = 1 + ^~2. 

1« 1 o o -, 3+V-55 -3+V-55. 

16. x=l,y=—2;x=2,y=—l;x= 1 ,y= j ; 

or,x = , y = y. . 

17 9 o o o -1+3V^3 1+3^. 

17. x=2,y = 3;a:=-3,y=-2;a: = ^ ,y= J , 

-1-3 y/^3 1-3 V / ' =r 3 
or, a;= ^ » ? = ^ ■ 

18 . , = 3 ;y =2;, = 2 ,,=3 i ,=^±^,,==^. 9 ; 

_-9-v/309 -9 + ^309 



12 " ,y ~ 12 

19. 3=1, y=-3; 3=-l, y=3; 3=141, y=3f; or, 3=-14f, 

20. 3 = 2,y = 3; or,3 = 2f>y = lf. 

01,100 4 22 59 

21. a: = 4, y = 2, « = 3; or, x = -,ij = — , z = -g-- 

22. a; = 1, y = 2, z = 4 ; a; = — 1, y = — 2, s = — 4 ; x — 9, 

y = — 6, s = 4 ; or, x = — 9, y = 6, z = — 4. 

Art. 322; pages 246 to 248. 

4. 12 and 7, or — 12 and — 7. 5. 11 and 7, or — 11 and — 7. 

6. A, $2025; B, $900; or, A, $900; B, $2025. 



ANSWERS TO EXAMPLES. 445 

7. A, 25 ; B, 30. 8. Length, 150 yds. ; breadth, 100 yds. 

9. 13 and 6. 10. A, $15; B, $80. 11. 10 lbs., at 8 cts. 
12. A, $ 5 ; B, $ 120. 13. Duck, $ 0.75 ; turkey, $ 1.25. 

14. Price, $ 1600 ; length, 1G0 rods ; breadth, 40 rods. 
15. Larger, 864 sq. in. ; smaller, 384. 16. A, $ 275 ; B, $ 225. 
17. 1st rate, 7 p.c; 2d, 6. 18. A, 40 acres at $ 8 ; B, 64, at $ 5. 

19. Distance of towns, 450 miles ; A, 30 miles a day ; B, 25. 

20. 3 and 1 ; or, 2 + sj 7 and 2 - y/ 7. 21. Larger, 12 ft. ; 
smaller, 9. 22. Width of street, 63 ft.; length of ladder, 45. 

23. B, 15 days ; C, 18 days. 
24. Length, 16 yds. ; width, 2 yds. 

Art. 328 ; page 253. 

3. (a: + 60) (a; + 13). 6. (x + 13) (a? -3). 9. (4a;-l) (2a; + 5). 

4. (x -9) (x -2). 7. (jc-5)(2a; + 3). 10. (x - 3) (4 x - 3). 

5. (a; -10) (a; + 6). 8. (7 a; + 3) (3 a; + 7). 11. (a; + 2) (2 a; -3). 

12. (3a:-2 + v/3)(3a-^2- v /3). 13. (y/17 + 4 + a-) (^17-4-3:). 
14. (7a3 + l + 2 v / 5)(7a; + l-2v/5). 

Art. 329 ; page 254. 

2. ar + a; = 2. 5. 3 x 2 - 2 x = 133. 8. 3 x 1 + 17 x = 0. 

3. a; 2 -9 a; =-20. 6. 21.x 2 + 44 x = 32. 9. a: 2 -2a; = 4. 



4. 5a; 2 - 12a; = 


9. 7. 6ar+35a;=-49. 10. 
Art. 330 ; page 255. 


x 2 — 2 m x=n— m 4 


7. 0or y . 


8. or — 4. 9. or ± 3. 


in 5 x 


lL-*orl 

a c 


12. ± 2 or ± 3. 


1 5 

13 --3° r± 2 


14. ±sja 


or — . 15. 0, - 


5 7 1 
■2'3'° r ~4- 



16. 2,3,-3, -4, I, or -5. 



446 ALGEBRA. 

Art. 331; page 256. 



3. (x + y/2 x + 1) (x - \/2 x + 1). 

4. (a; + \Jx + 1) (x — \J x + 1). 

5. (a + ^5lTb + b) (a - \j~h~ab + b). 

6. (x°~ + 3 x y + f) (x 2 -3xy + f). 

7. (x + 1 + S/Sx + 2) (x + 1 - \J~3x~+2). 

8. (m 2 + mn + w 2 ) (m 2 — m n + to 2 ). 



Art. 332 ; page 256. 



2. 






7. ^ 7 * V 71 - 1 ™ -v / 7±V /: -l 



or 



2^2 2^2 

Art. 357; pages 269 and 270. 

1. 4. 2. 11. 3. £. 4. 1 J. 5. ±4. 6. ± 12. 

7. ± 14. 8. 25 and 20. 9. 23 and 27. 10. 12 and 15. 
11. 8 and 18. 12. 26 and 14. 13. 17 and 12. 14. 12 and 8. 
15. First, 1:2; second, 2 : 1. 16. Females : males = 4:5. 

17. 8 : 7. 

Art. 365; page 273. 

2.4. 3. „ = 8* 4* 5.4. 6. y= 14 . . 

°> 4 — 5cc 

7. 10 inches. 8. 3 (y/ 2 — 1) inches. 9.143. 



ANSWERS TO EXAMPLES. 447 



Art. 370; page 276. 

3. 1=71, £=540. 4. Z=-69, £=-620. 5. 1=57, £=552. 

OQ A9 3 

6. l=-U5, £=-2175. 7. Z=^|,£=y. 8. l=--,S=0. 

9. l=-^,S=±. 10. l=~,S=~. U.l=5,S=17. 

4 2 



Art. 371; pages 278 and 279. 

q 95 1 

4. a = 3,£ = 741. 5. a = ^,i==--^-. 6. <Z=-,£=39. 

7. d = — i, « = -?. 8. a=o,d=-3. 9. w=18, £=411. 
12 4 

10. «Z = — 8, n = ll. 11. w = 30, Z = 80. 12. ^ = 52, a = 4; 
or, » = 43, a — — 5. 13. n = l%l = - 43. 



Art. 372; page 279. 

7 8 10 11 5 3 1 1 

A 3'3"I'"3"" J, 2' w, 2' '2' ' 2' 

4 2 3 _ 4 _ 5 5 Jl -^ _^ _!? -^ -?? 

*■ ^ — ^ 4 > °- 7 ' 7' 7' 7' 7' 7 

2 6 14 22 „ am + 5 a(m -l) + 2b 



5'5' 5 ' 5 ' ' m + 1 ' m + 1 

Art. 373; page 281. 
3. 2500. 4. Last payment, $ 103 ; amount, $ 2704. 

5. 4. 6. After 9 days, at a distance of 90 leagues. 

7. 4, 11, 18, and 25. 8. 3. 9. 0. 10. 20 miles. 

11. 2, 6, 10, and 14 ; or, - 2, - 6, - 10, and - 14. 12. 8. 



448 ALGEBRA. 

Art. 378; page 284. 

4. 1 = 2048, £=4095. 9. 1 = - — , s = -^^ 

64 ' 192 

64 2059 _ 1 511 

5- l = m> S= 243-. 10. 1 = ^, S=~. 

6. Z = 2048, £=1638. 11. l^-~, £=^. 

7 l--- L s- 3 ^ 12 Z- 1 <?- 341 

*~ 256' ^-256' "' ^-~768'^-~256- 

1 2047 

8 - l = zm> S =mi- 13^ = 192,5=129. 



Art. 379; page 286. 

- 1 c 341 2 _ 2 

4. a = ^, £ = 7 r-. 5. a = 7., Z = 



2' 2 "" "- _ 3' 6561' 

6. r = 3, £=2186; or, r = -3, £=1094. 

1 2457 

7. ^-j, /S= m . 8. » = 5, £=121. 

9. n = l, r = \. 10. w = 6, Z = -?S 1L w = 8, a = -l. 

^ 2 



Art. 380 ; pages 287 and 288. 



3. 4. 




5 3 




160 
y - 19" 


4 ? 

3 




ft 15 

Art. 381 


8 'i' 

; page 288. 


,0.-12. 


3.|. 4. 


13 
27" 


5 n 
5 - 15- 


ft 8G 7 

6 - 165* 7 ' 


17 237 
150* 1100 



ANSWERS TO EXAMPLES. 449 



Art. 382; pages 289 and 290. 

48 16 32 64 392781243 39 

'3' 9' 27' 81' 243" 2' 2' 2 ' 2 ' 2 ' ° r ' 2' 2' 

-|, |, -y. 5. - 6, -18, - 54, - 162, - 486, - 1458. 

927 81243 33333 3 3 

~4' 16' ""64' 256* 4' 8' 16' 32' 64' 128' 256' 

3 3 _3_ _3^ 3_ _3_ _3_ 
° r '~4' 8' ~16' 32' 64' 128' " 256' 



Art. 383; page 291. 

3. $ 64. 4. $ 295.23. 5. 3100 ft. 6. 5, 10, 20, and 40 ; 
or, - 15, 30, - 60, and 120. 7. - 4. 8. T V 

Art. 386 ; page 292. 

Z .E. 3. -i. 4. 3 " •» 



31' 78' ' 4' ' arc-&?i + 2&-a' 



2. 



Art. 387; page 293. 
48 24 16 12 48 8 48 



125' 65' 45' 35' 145' 25' 155' 



5_5_5 ■ 21 _7 21 21 

3 ' 4'" 3'" 2" '" '" 5 ' 3' 13' 17" 

_ (m + 1) a 6 (m + 1) a b (m + 1) ab 



m 



b + a ' m6 + 2a-6' mH3«-2i' 



Art. 397; pages 297 and 298. 
4. Of 4 letters, 360 ; of 3, 120 ; of 6, 720 ; in all, 1956. 
5. 1680. 6. 3838380. 7. 358800. 8. 15120. 9. 120. 
10. 35. 11. 15504. 12. 31824. 13. 77520. 14. 648. 



450 ALGEBRA. 

Art. 403 ; page 302. 

5. 1 + 5 c + 10 c 2 + 10 c 3 + 5 c 4 + c 5 . 

6. a 6 + 6 a 5 a 8 + 15 a 4 x 6 + 20 a 3 x 9 + 15 a?x u +Qa x lh + x ls . 

7. x s -8x 6 y + 24:X i 7f-S2x 2 f+16y\ 

8. a 7 ^-7a 6 i 6 c^ + 21a 5 i 5 cV 2 -35aH 4 c 8 ^ 3 +35aH 8 c 4 ^ 4 

-21a 2 & 2 c 5 d 5 +7a&c 6 a ,6 -c 7 cf. 

9. m 12 + 18 m 10 w 2 + 135 m 8 w 4 + 540 m 6 w 6 + 1215 m 4 w 8 

+ 1458 m- n 10 + 729 w 12 . 

10. a - 10 -20 a- 8 a^+160 ar 6 x- 640a~ ^+1280 or 2 ar-1024a^. 

11. c™ + 8 c*' cfl + 28 c 4 e^ + 56 c$~ <fi + 70 c% d 3 + 56 c 2 d^' 

+ 28 c* <# + 8 J eft + d 6 . 

12. m~^ + 14»"^ w 3 + 84 m- 8 » 6 +280 m - ^" w 9 + 560m~*» 12 

+ 672 m"^ w 15 + 448 j»~* w" 4- 128 w 21 . 

13. a- 4 -4ffl- 3 i 2 x^ + 6 a~ 2 Z> 4 a;^ - 4 or 1 6 6 a; + b s sA 

Art. 404; page 303. 

2. 5005 a 6 a; 9 . 4. - 19448 c 10 d 7 . 6. 42240 x~ 3 yK 

3.2002 m 6 . 5.495 a 8 . 7. 262440 a 2 ar 7 . 

Art. 405 ; page 304. 

2. 1 - 4 x + 2 a; 2 + 8 x 3 - 5 x i - 8 x h + 2 x 6 + 4 x 1 + x\ 

3. a; 6 + 9 x 5 + 30 a; 4 + 45 x 3 + 30 x 2 + 9 x + 1. 

4. l-6a; + 6a; 2 +16a: 8 — 12a 4 -24a; 8 -8a: 6 . 

5. l+5a; + 5a; 2 -10a; 3 -15a; 4 +lla; 5 +15x 6 -10a; 7 -5x 8 +5x 9 -cc 10 . 



Art. 414; page 309. 

3. l-2a: + 2a; 2 -2a: 3 + 2a; 4 

4. 3 + 19 x + 95 x 1 + 475 x a + 2375 a; 4 

5. 2-a; + 3a: 2 -a: 8 + 3a: 4 



ANSWERS TO EXAMPLES. 451 

6. l-2a; + 2a; 3 — 2x 4 + 2x 6 

7. l-2x + ox 2 -16x 3 + 4:7x i 

1 5x 7_x* lTjc 3 31_x 4 

9. 2- 7 a; + 28a; 2 - 91 a; 3 + 322 a; 4 

2 a; 7 a; 2 13 a; 3 8 a; 4 

io. i + — -— 27 + 81 

1 3jc a 2 15k 8 49a 4 
1L 2 + ~T + IT* 16 + 32 



Art. 415; page 310. 

2 a;- 2 4 a;- 1 _8_ 16 a; 32 a; 2 
2 - _ 3 _+_ 9 _+ 27 + 81 + 243 

3, £C - 1 + 3 + 2x-5a; 2 -16a; 3 

4. x~ 2 — x- 1 -2 x + 2 a; 2 -4 a; 3 



Art. 416; page 311. 

a; a; 2 a; 3 5 a; 4 a; 3 a; 2 3 X s 3 a; 4 

3 * 1 + 2~8~ + 16~128"" ,1+ 2 + 8 "16 + 128"' 

a; 2 a; 3 5 a; 4 _ a _ s 2 5 a; 3 10 a; 4 

3. l_aj_— — - — g-... «>. i 3 9 -~ 81 - 243 •'• 

x 4 „ . a; 2 a: 2 13 a; 3 8 a; 4 

4. l_a; + a; 2 + a; 3 + y ... 7. 1 + -+— — ^ +-^3 ... 



Art. 418 ; page 314. 

3 2.41 _J_ 6_ 

2 - x~T2 + x~=2' x-2 x 'x-7 x-6' 

, 3 2 ,1 J_ 7 _2 3_ 

«~T+8" & '^T4 + a; + l' '"2»-5 Sas + l" 

Q 1 2 _J 1 1 4 

8 "3+T£ + 3=!T y "6(x + l) 2(a;-l) i "3(a;-2)' 



452 ALGEBRA. 



Art. 419 ; page 316. 

o 11 1 „ 1 4 4 

2- — ^- + 7— ^Tv 2 + 7— "TV,- 4.-— + - — + 



x + l (x + l) 2 (x + l) 3 ' x-2 (x-2f ' (a; -2) 

3 2 3 5 3 6 



B' 



x-5 (x-5) 2 ' x + l (x + l)' 2 (x + l) 3 ' 

3 5 



6. 

7. 



2 (2 a; - 5) 2 (2 x - 5) 2 ' 
2 4 3 



3 a; + 2 (3x + 2) 2 (3x + 2) 3 



Art. 420 ; page 317. 
_ 2 3 5 .515 

2. — -, — ; . 4. 



x x+2 (x+2) 2 ' 'x x* x + l (x + l) 2 ' 

„11 1 1 .123 4 

3. -+— T + —0 + 7— 2K3- 5.- ,+ 



x x— 1 x—2 (x—2)' 2 ' ' x x 2 x s x + 5* 



■i 



x-2 2x-3 (2x-3) 
„ 5 1 2 5 4 

7. S + -T-— 



X X 2 X 3 X + l (x+l) 2 ' 

Art. 422; page 320. 

o 2,84 /i y 3?/ 19?/ 6 19 y' 

3. x = y — ?/ 2 + ?/ 8 — y 4 . . . 4. x = sL -j £ — . . 

J J J J 2 16 + 128 128 

5. x = y + y 3 + 2 y 5 + 5 y 7 . . . 

vy ; 2 i- 3 4 ■" 

7 x _ ?/ , ^,2^ 17 y' y 2y 2 y 3 14y* 

^~ y+ 3 + l5- + 315"- 8 ' *-3 + 27~~243~2T87" 



Art. 425; page 325. 

- § 5 | .15 1,5 a . 5 _a 
4. a 2 + -a 2 x + -^-a 2 x' + z—a 2 x z — — -a ^ 4 
«s 8 lb 128 



ANSWERS TO EXAMPLES. 453 



5. 1 - 6 x + 21 x 2 - 56 x 3 + 126 x* 



3 12 2 52 3 234 4 
b. i + 5 x + 25 x + 125 x + G25 x 

i 1 _i. 1 _| 1 _i 5 -z 

7. a J —-^a *x — -^a 'X' — zr^a 2 x 6 — zr^a J x* 

£ o lb lwo 

„ , 1 2 „ 14, 35 
8.1-- x + - x *-- x , + mX > 

9. ar 3 + 3a- 4 x + 6 a~ 5 x 2 + 10 a~ 6 x 3 + 15 «" 7 x* .... 
10. c"^— c- 3 df + c~* d 2 - c- 6 d 3 + c" 1 *' d* 



11. x z —2x b y — X s y 2 — -X s y 3 — ^x 3 y* 

o o 

I 3 15 7 35 3 a 315 jjl 6 

14. m + 6 in 6 n- + -jr- m J n 6 + -=- mr n 2 -\ — — m J ?i . . . 

2 2 o 

in -i -i^ i ™ o o 1760 „ „ 12320 

13. 1 — 10 xy- 1 + 80 a; 2 ?/- 2 q-b'jt'H 5— a? 4 ?/" 4 

2 2 8 

15. a 4 + 12 a 5 gT 2 + 90 a 6 y~ 4 + 540 a 7 jr 6 + 2835 a s y 



— 8 



Art. 426 ; page 326. 

33 aT'^'x'' 315 a 8 44 x^' y* 

"~2048 ' 128"' 6561 ' 



663 x "V 



4. 84 m 6 . 6. - " y ■ 8. 210 w^c" 8 . 

8192 

9. _?5? a -^aj- 6 . 10. 3§x- 3 «y- Vi z~ 1 £. 



Art. 427; page 327. 

3. 3.14138. 5. 9.94987. 7. 2.03054. 

4. 2.08008. 6. 1.96101. 8. 2.97183. 



454 



ALGEBRA. 



2. 
3. 



1 + x 



Art. 435; pages 331 and 332. 

4 -11 x „ 2 + 5« + 5a 2 



1 — x — x 2 ' 
a 



b + ex 



4. 
5. 



1 — 5 x + 6 x 2 

1+x 
1 - 2 x + x 2 ' 



6. 

7. 



(1 + a) 3 ' 
3 — x — 6 x 2 

l-2a;-a: 2 +2a; 3 



3. 3. 

8. 225. 



8. 



l + 2cc 
1 — x — x 2 



9. 



2 + 2 a; - 3 x 2 
l — x + x 2 — x & 



4. 



Art. 440; page 336. 
14. 5. 30. 6. 1365. 



7. 5050. 



9. 



n i + 2n z + n 2 
11. 165. 



10. 



6 n 6 + 15 m 4 + 10 w 8 



?« 



30 



12. 5525. 



3. 4.0514. 



1. 1.681241. 

2. 2.644438. 

3. 1.748188. 



Art. 443 ; pages 338 and 339. 

4. 3.634241. 5. 2.23830. 6. 44.24. 

7. $1,356. 



Art. 455 ; page 344. 

4. 1.991226. 7. 2.225309. 

5. 1.924279. 8. 3.848558. 

6. 2.753582. 9. 2.702430. 



10. 3.489536. 

11. 4.191785. 

12. 4158543. 



1. 1.176091. 

2. 2.096910. 

3. 0.154902. 



Art. 456 ; page 345. 

4. 2.243038. 7. 0.853872. 

5. 0.522879. 8. 1.066947. 

6. 1.045758. 9. 0.735954. 



Art 464; page 350. 

2. 8.724276-10. 4. 9.470704-10. 6. 1.527511. 

3. 1.714330. 5. 0.011739. 7. 8 780210-10. 



ANSWERS TO EXAMPLES. 



4jj 



8. 4.812917. 11. 9.942550-10. 14. 4.89381. 

9. 7.013150-10. 12. 3 863506. 15. 1.718451. 

10. 2.960116. 13. 8 640409-10. 16. 7.4984240-10. 

17. 9.275374-10. 18. 1.9792784. 



2. 76. 

3. .2954. 

4. 6.61005. 

5. 55606.5. 

6. .011089. 



Art. 465 ; page 352. 

7. 186 334. 12. .034277. 



8. .223905. 

9. 1000.06. 
10. 9.77667. 

• 11. .00130514. 
17. .00548803. 



18. 



13. 46.7929. 

14. 11.327. 

15. 8.63076. 

16. .2070207. 
734.9114. 



Art. 466; pages 353 and 354. 

1. 2.125240. 4. 3 108462. 7. 9.613158 - 10. 

2. 8.223962-10. 5. 9.594161-10. 8. 9.970036-10. 

3. 9.852169-10. 6. 7.315321-10. 9. 9.905232-10. 







Art. 468 ; pages 356 


1 to 358. 


1. 


.0341657. 


13. 1.70869. 


25. .580799. 


2. 


.650573. 


14. .788547. 


26. -.631188. 


3. 


13560.2. 


15. .680192. 


27. 83.5656. 


4. 


.136085. 


16. 2.24328. 


28. .297812. 


5. 


1.14720. 


• 17. .296850. 


29. 98.4295. 


6. 


1.41421. 


18. -.191680. 


30. 1.65900. 


7. 


1.49535. 


19. .644849. 


31. 3 07616. 


8. 


.0655264. 


20. .501126. 


32. .867674. 


9. 


-1.97221. 


21. 1.09872. 


33. -2.09389 


10. 


458.623. 


22. 1.06178. 


34. 46809.2. 


11. 


-.000113607. 23. 1.09328. 


35. .588142. 


12. 


5.88336. 


24. 1.65601. 


36. 1.80446. 




37. 


.00323011. 


38. .0334343. 



456 



ALGEBRA. 



The following are the values of the expressions in Art. 468, 
when calculated by seven-figure logarithms : 



1. 


.034165G8. 


13. 


1.708689. 


25. 


.58079S7. 


2. 


.0505727. 


14. 


.7885469. 


26. 


- .6311888 


3. 


13560 27. 


15. 


.6801947. 


27. 


83.56558. 


4. 


.1360851. 


16. 


2 243284. 


28. 


.2978123. 


5. 


1.147203. 


17. 


.2968501. 


29. 


98.42-J91. 


6. 


1.414214. 


18. 


- .1916795. 


30. 


1.658989. 


7. 


1 495349. 


19. 


.6443490. 


31. 


3.076162. 


8. 


.06552632. 


20. 


.5011282. 


32. 


.8676754. 


9. 


- 1.972211. 


21. 


1.098718. 


33. 


- 2.093891. 


10. 


458.5759. 


22. 


1.061780. 


34. 


46808.95. 


11. 


- .0001136063 


. 23. 


1.093280. 


35. 


.5881412. 


12. 


5 883366. 


24. 


1.656005. 


36. 


1.804459. 




37. .003230121 


38. 


,03343431. 






Art. 469 ; page 359 






3. , 


458156. 


5. 


- .494903. 


7. - 


-2.70951. 


4. . 


185339. 


6. 


- .260231. 


8. - 


- 10.2341. 



The results with seven-figure logarithms are as follows • 

3. .4581568. 5. - .4949028. 7. - 2.709513. 

4. .1853394. 6. -.2602272. 8. -10.23414. 



Art. 479 ; pages 368 and 369. 



1. 7. 


3. - 6. 5. 


7. 




7. 6. 


2. 6. 


4 .4 e. 


5. 




8. 7. 


9. 1.56937. 


13. 11.725 yrs. 




17. 


3.96913. 


10. 2.44958. 


14. $9756.59. 




18. 


7.18923. 


11. 2.00906. 


15. 7 per cent. 




19. 


- 2.4578. 


12. $5421.33. 


16. 9.392 yrs. 




20. 


- 1.07009 



ANSWERS TO EXAMPLES. 457 

The results of the last 12 examples, using seven-figure 
logarithms, are as follows : 

9. 1.569369. 13. 11.725 yrs. 17. 3.969124. 

10. 2.449576. 14. $9756.59. 18. 7.18922. 

11. 2.009056. 15. 7 per cent, 19. -2.457802. 

12. $5421.35. 16. 9.392 yrs. 20. -1.070092. 

Art. 489; page 373. 
2. 3 and - 5. 3. a and | (- 1 ± y/^3). 4. 2 and 2. 5. ± 4. 

6 . X 3_ 6a .2_ 6a ,_ 3 = 7> ? and 5 8 . ?and-l 

3 2 4 5 

Art. 490; page 374. 

2. z 3 + 9x 2 + 23£ + 15 = 0. 4. 6ai 3 -lla: 2 + 6*-l = 0. 

3. a 8 -19 a: -30 = 0. 5. cc 4 - 5x' + 4 = 0. 

6. a; 4 - 10 x* + 35 x 2 - 50 cc + 24 = 0. 

7. cc 3 - 13 x 2 + 56 x - 80 = 0. 

8. x 4 -6a; 3 +5x 2 + 12x = 0. 

9. 12 z 4 + 55 r 3 - 68 x 2 - 185 x + 150 = 0. 



Art. 494; page 375. 



5 



1. Sum, ; product, - 6. 2. Sum, - ; product, 12. 

3. 2±2v/2. 

Art. 504; page 382. 

2. ?/ 3 + 24?/ 2 + 191?/ + 498 = 0. 3. y* - 6?/ - if + 55y - 76 = 0. 



P 



■: 



Art. 505; page 383. 



2. y 2 -^ + <7 = 0. 4. ^-15y+26 = 0. 

y 3 + "2y ==0, 5 * 2/ 4 -6y 2 -137/-9 = 0. 



458 ALGEBRA. 

Art. 513; page 388. 

2. 1, 1, and 6. 4.-1,-1,-1, and 3. 

3. 2, 2, and 3. 5. 2, 2, 2, and - 6. 

Art. 517; page 390. 
2. - 1, 1, and 5. 3. 3. 4. 1. 5. 2. 

Art. 520; page 392. 
3. 1 + s/U. 4. 1 + V 15 - 5- - (1 + V^)- 6. - (1 + $5). 

Art. 527; page 399. 

3. Three ; respectively between and 1, 1 and 2, and — 1 and — 2. 

4. Three ; two between 1 and 2, and one between — 3 and — 4. 

5. One ; between 2 and 3. 

6. Four ; respectively between and 1, 1 and 2, 2 and 3, and 
— 2 and — 3. 

7. None. 

8. Two ; respectively between 2 and 3, and 3 and 4. 

Art. 532; page 403. 

3. — 1, - 2, and - 3. 9. A, and 1 ± }f—i. 

4. 2, -2, and -3. 

5. 2, 4, and - 1 ± y/^3. 

6. - , 4, and — ^ • 

7. 2, and — — *- . 

8. 3. 6, and - 2. 



10. 


1, 


2, and 3. 






11. 


3 

9 


, and ± 2 


v/- 


-2. 


12. 


o 








13. 


3. 








14. 


3, 


4. — 3, and ■ 


— 5. 



ANSWERS TO EXAMPLES. 459 

Art. 538; pages 407 and 408. 
2 1 9 ±x /77 3 ± y/5 l_^ ±V /^_ 2j? -3 

3. -1,1,1, or =^?. 6. 2,i,-3,or-|. 

4. ±l J±V /=T,or^f: 3 . 7. l,5,i or2±^3. 

y/33- 5 ±^42- 10 y/33 - y/33 - 5 ± y/42 + 10 y/33 

8 - 4 ' ° r ~I~ 

q _i 1 + y/ 5 ±V /2 V /5jri ~ Q " or 1-^5+^-2^5-10 
a> *' 4 4 

-l-^5±y^5-1 -l +v /5±y/-2v/5-10 
iu. w, 2~ ? 01 o 

Art. 541; page 410. 

3. -3or^±£3. 7. lorl±YE5. 

2 2_ 

4. 4 or 1 ± 4 \/=3. 8. 3 or 1 ± ^~ 3 ' 

5. 3, 3, or -2. 9. 2, 2, or— 1. 

6. 1,1, or -11. 10. $±-1(2. 

Art. 550; page 417. 
2.2.09455. 3.7.61728. 4. 1.3569, 1.6920, and -3.0489. 
5. 14.95407. 6. 2.2674 and 36796. 

7. 2.85808,-60602, .44328, and - 3.90738. 

Art, 551; page 419. 

2. 3.864854. 4. 2.4257. 6. 10.2609. 

3. 4.11799. 5. .66437. 7. 8.414455. 

Art. 552; page 420. 
2. 153209. 3. 1.02804. 



T A B L E 



CONTAINING THE 



LOGARITHMS OF NUMBERS 



FROM 1 TO 10,000. 



No. 


Log. 


No. 


Log. 


No. 


Lo£. 


No. Log. 


No. 


Loft. 


1 


0.000000 


21 


1.322219 


41 


1.612784 


61 


1.785330 


81 


1.908485 


2 


0.30103C 


22 


1.342423 


42 


1.623249 


62 


1.792392 


82 


1.913814 


3 


0.477121 


23 


1.361728 


43 


1.633468 


63 


1.799341 


83 


1.919078 


4 


0. CO 20 GO 


24 


1.380211 


44 


1.643453 


64 


1.806180 


84 


1.924279 


6 


0.698970 


25 


1.397940 


45 


1.653213 


65 


1.812913 


85 


1.929419 


6 


0.778151 


26 


1.414973 


46 


1.662758 


66 


1.819544 


86 


1.934498 


7 


0.845098 


27 


1.431364 


47 


1.672098 


67 


1.826075 


87 


1.939519 


8 


0.903090 


28 


1.447158 


48 


1.681241 


68 i 1.832509 


88 


1.944483 


9 


0.954243 


29 


1.462398 


49 


1.690196 


69 


1.838849 


89 


1.949390 


10 


1.000000 


30 


1.477121 


50 


1.698970 


70 


1.845098 


90 


1.954243 


11 


1.041393 


31 


1.491362 


51 


1.707570 


71 


1.851258 


91 


1.959041 


12 


1.079181 


32 


1.505150 


62 


1.716003 


72 


1.857332 


92 


1.963788 


13 


1.113943 


33 


1.518514 


53 


1.724276 


73 


1.863323 


93 


1 .968483 


14 


1.146128 


34 


1.531479 


54 


1.732394 


74 


1.869232 


94 


1.973128 


15 


1.17C091 


35 


1.544068 


55 


1.740363 


75 

76 


1.875061 


95 


1.977724 


16 


1.204120 


36 


1.556303 


56 


1.748188 


1.880814 


96 


1.982271 


17 


1.230449 


37 


1.568202 


57 


1.755875 


77 


1.88641)1 


97 


1.986772 


18 1.255273 


38 


1.579784 


58 


1.763428 


78 


1.892095 


98 


1.991226 


19 I 1.278754 


39 


1.591065 


59 


1.770852 


79 


1.897627 


99 


1.995635 


20 | 1.301030 


40 


1 .602060 


60 


1.778151 


80 1.903090 


too 


2.000000 



LOGARITHMS 



I 2 | 3 | 



8 | 9 



N.| 



I 1 



D. 



100,001)000 



4321 
8000 

012837 
7033 

021189 
6300 
9384 

0334 21 
7420 



000434 
4751 
9020 

013259 
7451 

021003 
6715 
9789 

033820 
7825 



110 
1 
2 

3 

4 
6 


7 
S 
9 



041393 
6323 
9218 

053078, 
0905 

0G0098 
4458 
8180 

071882 
6547 



041787 
5714 
9000 

053403 
7280 

001075 
4832 
8557 

072250 
5912 



000808 
6181 
9451 

013080 
7808 

022010 
0125 

030195 
4227 
8223 



001301 
5009 
9876 

014100 
8284 

022428 
0533 

030000 
4028 
8020 



001734 
6038 

010300 
4521 
87CT, 

02284 1 
0942 

031004 
5029 
9017 



042182 
0105 
9993 

053840 
7066 

001452 
6200 
8928 

072017 
0270 



042570 
6495 

050380 
4230 
8040 

061829 
5580 
9298 

072985 
6640 



042909 
0885 

0507C6 

, 4013 
8426 

002200 
5953 
9008 

073352 
7004 



002100 
6406 

010724 
4940 
9110 

023252 
7350 

031408 
5430 
9414 



002598 

6894 

011147 

5300 

9532 

023004 

7757 

031812 

6830 

9811 



003029 
7321 

011570 
5779 
9947 

024075 
8164 

032216 
6230 

040207 



003401 
7748 

011993 
6197 

020301 
4480 
8571 

032619 
6629 

040002 



003891 432 
8174428 

012415'424 
00l0i420 

020775416 
4896 412 
8978408 



033021 

7028 

040998 



404 
400 
397 



120 
1 

2 
3 



079181 

082785 

6360 

9905 
4 093422 
5, 6910 
6 100371 
7- 3S04 
8' 7210 
91 10590 



079543 

083144 
6710 

090258 
3772 
7257 

100715 
4146 
7549 

110926 



079904 

083503 
7071 

09001 i 
4122 
7004 

101059 
4487 
7888 

111203 



080206 
3801 
7420 

090903 
4471 
7951 

101403 
4828 
8227 

111599 



0801.20 


080987 


4219 


4576 


7781 


8136 


091315 


091007 


4820 


6109 


8298 


8644 


101747 


102091 


6109 


5510 


8505 


8903 


111934 


112270 



1149441 
8265 

121560! 
4830J 
8076J 

131298 
4496J 
7071 

140822! 
3951 



043302 
7275 
f 51153 

4990 
8805 

002582 
6326 

070038 
3718 
7308 



043755 

7664 
051538 
5378 
9185 
062958 
6699 
070407 
4085 
7731 



044148 
8053 

051924 
6760 
9563 

003333 
7071 

070770 
4451 
8094 



081347 
4934 
8490 

092018 
5518 
8990 

102434 
6851 
9241 

112005 



044540 
8442 

052309 
0142 
9942 

063709 
7443 

071145 
4816 
8457 



081707 
6291 
8845 

092370 
6866 
9335 

102777 
6191 
9579 

112340 



082007 
5047 
9198 

092721 
6215 
9681 

103119 
6531 
9916 

113275 



044932393 
88301390 



052694 

6524 
060320 

4083 

7815 
0716141370 

5182366 

8819363 



380 
383 
379 
376 
373 



0824201360 
6004 



9552 
093071 

6502 
100020 

3462 

6871 
110253 

3 CO 9 



357 
355 
352 
349 
346 
343 
341 
338 
335 



130,113943 


1 


7271 


2 


120574 


3 


3852 


4 


7105 


5 


130334 


6 


3539 


7 


6721 


8 


9879 


9 


143015 



114277 
7003 

120903 
4178 
7429 

130055 
3858 
7037 

140194 
3327 



114011 
7934 

121231 
4504 
7753 

130977 
4177 
7354 

140508 
3039 



115278 
8595 

121888 
5156 
8399 

131019 
4814 
7987 

141136 
4203 



115011 
8926 

122216 
5481 
8722 

131939 
5133 
8303 

141450 
4574 



115943 
9256 

122544 
5800 
9045 

132200 
5451 
8018 

141703 
4885 



116276 
9586 

122871 
6131 
9308 

132580 
5769 
8934 

142070 
5190 



116608 
9915 

123198 
6456 
9090 

132900 
6080 
9249 

142389 
5507 



116940333 
1-20245 330 
3525 328 
6781325 
130012 323 
3219 321 
6403i318 



9504 

142702 

5818 



310 
314 
311 



140 
11 
2 
3 
4 
5 
6 
7 
8 
9 

150 
1 
2 
3 
4 
5 
6 
7 
8 
9 

NTT 



140128 
9219 

152288 
6330 
8302 

101308 
4353 
7317 

170202 
3180 

176091 
8977 

181844 
4091 

7521 
190332 

3125 
5900 
8057 
201397 




146438 
9527 

152594 
5640 
8664 

101007 
4050 
7013 

170555 
3478 

176381 

9204 

182129 

4975 

7803 

190012 

3403 

0170 

8932 

201070 

i 



140748 
9835 

152900 
5.U3 
8005 

101907 
4947 
7908 

170848 
3709 



,147058 

150142 
3205 
0240 
9206 

1 02200 
5244 
8203 

171141 
4000, 



170070 
9552 

182415 
62591 
8084 1 

190892 

31.81 

0153 

9206 

201913 



176959 
9839 

182700 
6542 
8306 

191171 
3959 
0729 
9481 

202216 

3 I 



147307 

150449 
3510 
0549 
9507 

102504 
5541 
8497 

171434 
4351 

177248 

180120 
2985 
6825 
8047 

191451 
4 237 
7005 
9755 

2(1 2 1 88 



147070 

'S0750 
3815 
3852 
9808 

102803 
5838 
8792 

171720 
4641 

177530 

180413 
3270 
0108 
8928 

191730 
4514 
7281 

200029 
2761 



II 



147985 

151003 
4120 
7154 

160108 
3101 
6134 
9086 

172019 
4H32 

177825 

180099 
3555 
639J 
9209 

192010 
4792 
7556 

200303 
3033 
o 



148291 

151370 
4424 
7457 

100409 
3400 
6430 
9380 

172311 
5222 

178113 

1801180 
3839 
6674 
9490 

19228H 
6069 
7832 

1200577 
3305 



148003 

151670 
4728 
7759 

100709 
3758 
0720 
9074 

172003 
6512 



148911 

151982 
5032 
f-001 

161008 
4055 
7022 
9908 

172895 
6802 



309 

307 

305 

303, 

30ll 

299- 

297i 

295 

293 

291 



178401 

181272 
4123 
6956 
9771 

192567 
6346 
8107 

200850 
3577 



178689 289 

181558 287 
4407 285 
7239.283 

190051 '281 
2840 279 
8623 278 
8382276 

201124 271 
3818 272 
T~Ti | n 



OF NUMBERS. 



N.| | ] 2 3 


4 


5 


6 7 8 9 ,D. 


160 204120 204391 


204663 


204934 


205204 


205475 


205746 


206016 206286 


206556)271 


1 


G82G 


709G 


73G5 


7G34 


7904 


8173 


8441 


8710 


8979 


9247 269 


2 


9515 


9783 


210051 


210319 


210586 


210853 


211121 


211388 


211654 


211921'2C7 


3 


212188 212454 


2720 


2986 


3252 


3518 


3783 


4049 


4314 


4579 266 


4 


4844 


6109 


6373 


6638 


5902 


6166 


6430 


6694 


6957 


7221 


264 


5 


7484 


7747 


8010 


8273 


8536 


8798 


9060 


9323 


9585 


9846 


262 


6 


220108 


220370 


220631 


220892 


221153 


221414 


221675 221936 


222196 


222456 


261 


7 


2716 


297G 


3236 


3496 


3755 


4015 


4274 


4533 


4792 


6051 


259 


8 


5309 


5568 


5826 


6084 


6342 


6600 


6858 


7115 


7372 


7630 


258 


9 


7887 


8144 


8400 


8057 

231215 


8913 
231470 


9170 

231724 


9426 
23T9~79 


9G82 
232234 


9938 230193 


256 


170 


230449 230704 


230960 


232488 232742 


255 


1 


2996 3250 


3504 


3757 


4011 


4264 


4517 


4770 


6023 


6276 


253 


2 


5528 5781 


6033 


6285 


6537 


6789 


7041 


7292 


7544 


7795 


252 


3 


804G 1 8297 


8548 


8799 


9049 


9299 


9550 


9800 


240050 


240300 


250 


4 


240549 240799 


241048 


241297 


241546 


241795 242044 


242293 


2541 


2790 


249 


6 


3038 


3286 


3534 


3782 


4030 


4277 


4525 


4772 


5019 


6266 


248 


6 


5513 


5759 


6006 


6252 


6499 


6745 


6991 


7237 


7482 


7728 


246 


7 


7973 


8219 


84G4 


8709 


8954 


9198 


9443 


9687 


9932 


250176 


•245 


8 


250420 


250GG4 


250908 


251151 251395 


251638 


251881 


252125 


252368 


2610 


24o 


9 


2853 


3096 
255514 


3338 
255755 


3580| 3822 


4064 


4306 
256718 


4548 


4790 


5031 


242 


180 


255273 


255996 


256237 


256477 


256958 


257198 


257439 


241 


1 


7679 


7918 


8158 


8398 


8637 


8877 


9116 


9355 


9594 


9833 


239 


2 


260071 


260310 


260548 


260787 


261025 


261263 


261501 


261739 


261976 


262214 


238 


3 


2451 


2G88 


2925 


3162 


3399 


3G3G 


3873 


4109 


4346 


4582 


237 


4 


4818 


5054 


6290 


5525 


6761 


6996 


6232 


6467 


6702 


G037 


235 


6 


7172 


7406 


7641 


7875 


8110 


8344 


8578 


8812 


904G 


9279 


234 


6 


9513 


9746 


9980 


270213 


270446 


270679 


270912 


271144 


271377 


271609 


233 


1 


271842 


272074 


272306 


2538 


2770 


3001 


3233 


3464 


3696 


3927 


232 


8 


4158 


4389 


4620 


4850 


5081 


5311 


6542 


6772 


6002 


6232 


230 


9 

190 


6462 

278754 


6692 


6921 


7151 


7380 


7609 


7838 


8067 


8296 


8525 


229 


278982:279211 


2794391279667, 

281715281942 


279895 28 


280351 21 


228 


1 


281033 


281261:281488 


282169 


2396 


2622 


2849 


3075 


227 


2 


3301 


3527 3753 


3979 4205 


4431 


4656 


4882 


5107 


5332 


226 


3 


6557 


5782! G007 


6232 


6456 


6681 


6905 


7130 


7354 


7578,225 


4 


7802 


8026 8249 


8473 


8696 


8920 


9143 


93G6 


9589 


9812 223 


5 


290035 


2902571290480 


290702 


290925 


291147 


291369 


291591 


291813 


292034 


222 


6 


2256 


2478 2699 


2920 


3141 


3363 


3584 


3804 


4025 


4246 


221 


7 


4466 


4687 4907 


5127 


5347 


5567 


5787 


6007 


6226 


6446 


220 


8 


6665 


6884 


7104 


7323 


7542 


7761 


7979 


8198 


8416 


8G35 


219 


9 


8853 


9071 


9289 


9507 


9725 


9943 


3001G1 


300378 300595 300813 


218 


200 301030 


301247 


3014G4 


30168130 


302114 


302331 


302547 


302764 302980:217 


1 


319G 


3412 


3628 


3844 


4059 


4275 


4491 


4706 


4921 


5136 216 


2 


5351 


5566 


5781 


5996 


6211 


6425 


6639 


6854 


7068 


7282 215 


3 


749G 


7710 


7924 


8137 


8351 


8564 


8778 


8991 


9204 


9417 


213 


4 


9G30 


9843 


310056 


310268 


310481 


310693 


310906 


311118 


311330 


311542 


212 


5 


311754 


311966 


2177 


2389 


2G00 


2812 


3023 


3234 3445 


3656 


211 


6 


38G7 


4078 


4289 


4499 


4710 


4920 


5130 


6340 


5551 


5760 


210 


7 


5970 


6180 


6390 


6599 


6809 


7018 


7227 


7436 


7646 


7854 


209 


8 


80G3 


8272 


8481 


8G89 


8898 


9106 


9314 9522 


9730 


9938 


208 


9 320146 320354 3205G2|320709 


320977 321184 


321391)321598 321805 


322012 207 


210 


322219 


3224261322633 


322839 


32304G 


323252 


3234581323665 


323871 


324077 


206 


I 


4282 


4488 4G94 


4899 


5105 


5310 


6516 5721 


5926 


6131 


205 


2 


6336 


6541 G745 


6950 


7155 


7359 


7563 7767 


7972 


8176 


204 


3 


8380 


8583 8787 


8991 


9194 


9398 


9601 9805 330008 


330211 


203 


4 


330414 


330617 330819 


3?1022 


331225 


331427 


331630331832 


2034 


2236 


202 


5 


2438 


2640, 2842 


3044 


3246 


3447 


3649, 3850 


4051 


4253 


202 


6 


4454 


4655| 4856 


5057 


5257 


5458 


5658 5859 


6059 


6260 201 


7, 6460 


6660, G8G0 


70(0 7260 


7459 


7659 7858 8058 


8257 


200 


8 845G 


8650 3855 


90541 9253 


9451 


9G50 1 98493400471340246 


199 


9 , 340444 340642 l 340841 


3410391341237 


341435 


341G323418301 20281 22251138 


N.| 


1 


2 


3 | 4 |, 


i) 7 | 8 


■9 


D. 



LOGARITHMS 



N. | 1 


2 


3 | 4 5 


6 


7 | 8 


9 ID 


220 342423 


342020 


312817 


343014 


343212 


343409 


343000 


343&02 


343999 


344196 


197 


1 


4392 


4589 


4785 


4981 


6178 


5374 


6570 


5766 


6962 


6167 


196 


2 


6353 


6549 


6744 


6939 


7135 


7330 


7525 


7720 


7915 


8110 


195 


3 


8305 


8500 


8694 


8889 


9083 


9278 


9472 


9666 


9860 


350054 


194 


4 


350248 


350442 


350636 


350829 


351G23 


351216 


351410 


351(03 


351796 


1989 


193 


6 


2183 


2375 


2568 


2761 


2954 


3147 


3339 


3532 


3724 


3916 


103 


6 


4108 


4301 


4493 


4685 


4876 


6068 


5260 


6452 


6643 


5834 


102 


7 


6026 


6217 


6408 


6599 


6790 


6981 


7172 


7363 


7554 


7744 


101 


8 


7935 


8125 


8316 


8506 


8696 


8886 


9076 


9266 


9456 


9646 


190 


9 


9835 


360025 


360215 360404 


360593 


360783 


360972 


361161 


361350 


361539 


189 


230 


361728 


361917 


362105 


362294 


362482 


362671 


362859 


363048 


363236 363424 


188 


1 


3612 


3800 


3988 


4176 


4363 


4551 


4739 


4926 


5113 


5301 


188 


2 


6488 


6675 


5862 


6049 


6236 


6423 


6610 


6796 


6983 


7169 


187 


3 


7356 


7542 


7729 


7915 


8101 


8287 


8473 


8659 


8845 


9030 


180 


4 


9216 


9401 


9587 


9772 


9958 


370143 


370328 


370513 


370098 


370883 


185 


5 


371068 


371253 


371437 


371622 


371806 


1991 


2175 


2360 


2544 


2728 


184 


6 


2912 


3096 


3280 


3464 


3647 


3831 


4015 


4198 


4382 


4565 


184 


7 


4748 


4932 


6115 


6298 


5481 


5664 


5846 


6029 


6212 


6394 


183 


8 


6577 


6759 


6942 


7124 


7306 


7488 


7670 


7852 


8034 


8216 


182 


9 8398 


8580 


8761 


8943 


9124 


9306 


9487] 9068 


9849 


380030 


181 


240 


380211 


380392 


380573 


380754 


380934 


381115381296 


381476 


381056 


381837 


181 


1 


2017 


2197 


2377 


2557' 2737 


2917 3097 


3277 


3456 


3636 


180 


2 


3815 


3995 


4174 


4353! 4533 


4712 4891 


6070 


5249 


5428 


179 


3 


5606 


6785 


6964 


6142 


6321 


6499 


0077 


6856 


7034 


7212 


178 


4 


7390 


7568 


7746 


7923 


8101 


8279 


8456 


8634 


8811 


8989 


178 


5 


9166 


9343 


9520 


9698 


9875 


300051 


390228 


390405 


390582 390759 


177 


6 


390935 


391112 


391288 


391464 


301641 


1817 


1993 


2109 


23451 2521 


176 


7 


2697 


2873 


3048 


3224 


3400 


3575 


3751 


3926 


4101 4277 


176 


8 


4452 


4627 


4802 


4977 


5152 


5326 


5501 


6676 


5850 6025 


175 


9 


6199 


6374 


6548 


6722 


6896 


7071 


7245 


7419 


7592| 7766 


174 


250 


397940 


398114 


398287 


398461 398634 


308808 398981,399154 


399328 399501 


173 


1 


9674 


9847 


400020 


400192 400365 


400538 4007111400883 


401056 


401228 173 


2 


401401 


401573 


1745 


1917 


2089 


2261 


24331 2605 


2777 


2949 


172 


3 


3121 


3292 


3464 3635 


3807 


3978 


4149 4320 


4492 


4663 


171 


4 


4834 


6005 


6176 


5346 


6517 


6688 


5858 


6029 


6199 


6370 


171 


5 


6540 


6710 


6881 


7051 


72L1 


7301 


7561 


7731 


7901 


8070 170 


6 


8240 


8410 


8579 


8749 


8918 


9087 


9257 


9426 


9595 


9764 It. 9 


7 


9933 


410102 


410271 


410440 


410609 


410777 


410940 


411114 


411283 


411451,169 


■ 8 


411620 


1788 


1956 


2124 


2293 


2461 


2629 


2796 


2964 


3132168 


9 

260 


3300 

414973 


3467 


3635 


3803 


31)70 


4137 


4305 


4472| 4639 


4806|l67 


415140 


415307 


415474 


415641 


415808 


415974 


416141 410308,4: 


1 


6641 


6807 


6973 


7139 


7306 


7472 


7638 


7804 


7970! 8135|loo 


2 


8301 


8467 


8633 


8798 


8964 


9129 


9295 


9400 


962.'. 97911 10.) 


3 


9956 


420121 


420286 


420451 


420616 


420781 


420045 


421110 


421275 


421439 l«o 


4 


421604 


1768 


1933 


2097 


2261 


2426 


2590 


2754 


2918 


3082 


104 


5 


3246 


3410 


3574 


3737 


3901 


4065 


4228 


4392 


4555 


4718 


164 


6 


4882 


6045 


5208 


6371 


6534 


5697 


5860 


6023 


6186 


6349 


163 


7 


6511 


6674 


6836 


6999 


7161 


7324 


7486 


7648 


7811 


7973 


162 


8 


8135 


8297 


8459 


8621 


8783 


8944 


9106 


9268 


0120 


9591 


162 


9 


9752 


9914 


430075 430236 


430308 


430559 


430720 


430881 


431042 


431203 


Hil 


270 


431364 


431525 


431685 


431846 


432007 


432167 


432328 


432488 


432640 432809 


161 


1 


2969 


3130 


3290 


3450 


3610 


3770 


3930 


4090 


4249i 4409 100 


2 


4569 


4729 4888 


6048 


6207 


6367 


6526 


6685 


6844 


6004 159 


3 6163 


6322 6481 


6640 


6799 


6957 


7116 7275 


7433 


75921159 


4 


7751 


7909! 8067 


8226 


83X4 


8542 


870l] 8859 9017 


9175ll58 


6 


9333 


9491 9648 


9806 


9964 


440122 440279440137 


440594 


440752; 158 


6 4 40909 44106(14 11224 


441381 


441538 


1605 


1852 1 2009 


2106 


2323 157 


7 


2480 


2637 


2793 


2950 


3100 


3203 


3419 


3576 


3732 


3889 167 


8 


4045 


4201 


4357 


4513 


4669 


4825 


4981 


5137 


6293 


5449 


156 


9; 6604 


6760 


5915 


6071 


6226 


6382 


6537 


6692 


6848 7003 


165 


N | 0' 


1 


3 9 | D. 



OF NUMBERS. 



N.I 


1 | 2 


3 | 4 


5 | 6 | 7 


8 | 9 | 


D. 




280 


447158 


447313 


447408 


447023 ( 


147778 


447933 448088 


448242 


4483 9' 7 


148552 


15.. 




1 


870C 


8861 


9015 


9170 


9324 


9478 


9633 


9787 


9941 


450095 


154 




2 


450249 


450403 


450557 


450711 ' 


150865 


451018 


451172 


461326 


451479 


1633 


154 




3 


1786 


1940 


2093 


2247 


2400 


2553 


2706 


2859 


3012 


3165 


153 




4 


3318 


3471 


3624 


3777 


3930 


4082 


4235 


4387 


4540 


4692 153 




5 


4845 


4997 


5150 


6302 


5454 


5606 


5758 


6910 


6062 


6214 


152 




6 


6366 


6518 


6670 


6821 


6973 


7125 


7276 


7428 


7579 


7731 


152 




7 


7882 


8033 


8184 


8336 


8487 


8638 


8789 


8940 


9091 


9242 


151 




8 


9392 


9543 


9694 


9845 


9995 


460146 


460296 


460447 


460597 


460748 


151 




9 


460898 


461048 


461198 


461348 


161499 


1649| 1799 


1948 


2098 


2248 


150 




290 


462398 


462548 


462697 


462847 - 


162997 


463146 


463296 


463445 


463594 


463744 150 




1 


3893 


4042 


4191 


4340 


4490 


4639 


4788 


4936 


5085 


6234 149 




2 


6383 


6532 


6680 


5829 


6977 


6126 


6274 


6423 


6571 


6719 149 




3 


6868 


7016 


7164 


7312 


7460 


7608 


7756 


7904 


8052 


8200 148 




4 


8347 


8495 


8643 


8790 


8938 


9085 


9233 


9380 


9527 9675 148 




5 


9822 


9969 


470116 


470263 - 


170410 


470557 


470704 


470851 


470998 


471145 147 




6 


471292 


471438 


1685 


1732 


1878 


2025 


2171 


2318 


2464 


2610 


146 




7 


2756 


2903 


3049 


3195 


3341 


3487 


3633 


3779 


3925 


4071 


146 




8 


4216 


4362 


4508 


4653 


4799 


4944 


5090 


5235 


6381 


6526 


146 




9 


5671 


5816 


5962 


6107 


6252 


6397 


6542 


6687 


6832 


6976 


145 




300 


477121 


477266 


477411 


477555 


477700 


477844 


477989 


478133 


478278 


478422.14 




1 


8566 


8711 


8855 


8999 


9143 


9287 


9431 


9575 


9719 


9863 


144 




2 


480007 


480151 


480294 


480438 


480582 


480725 


480869 


481012 


481156 


481299 


144 




3 


1143 


1586 


1729 


1872 


2016 


2159 


2302 


2445 


2588 


2731 


143 




4 


2874 


3016 


3159 


3302 


3445 


3687 


3730 


3872 


4015 


4157 


143 




5 


4300 


4442 


4585 


4727 


4869 


5011 


6153 


5295 


6437 


6579 


142 




6 


6721 


6863 


6005 


6147 


6289 


6430 


6572 


6714 


6855 


6997 


142 




7 


7138 


7280 


7421 


7563 


7704 


7845 


7986 


8127 


8269 


8410 


141 




8 
9 


8551 
9958 


8692 
490099 


8833 
490239 


8974 
490380 


9114 
490520 


9255 


9396 


9537 
490941 


9677 
491081 


9818 
491222 


141 
140 




490661 


490801 




310 


491362 


491502 


491642 


491782 


491922 


492062 


492201 


492341 


492481 


492621 


140 




1 


2760 


2900 


3040 


3179 


3319 


3458 


3597 


3737 


3876 


4015 


139 




2 


4155 


4294 


4433 


4572 


4711 


4850 


4989 


6128 


5267 


5406 


139 




3 


5544 


5683 


6822 


6960 


6099 


6238 


6376 


6515 


6653 


6791 


139 




4 


6930 


7068 


7206 


7344 


7483 


7621 


7759 


7897 


8035 


8173 


138 




5 


8311 


8448 


8586 


8724 


8862 


8999 


9137 


9275 


9412 


9550 


138 




6 


9687 


9824 


9962 


500099 


500236 


500374 


500511 


500648 


500785 


500922 


137 




7 


501059 


501196 


501333 


1470 


1607 


1744 


1880 


2017 


2154 


2291 


137 




8 


2427 


2564 


2700 


2837 


2973 


3109 


3246 


3382 


3518 


3655 


136 




9 


3791 


3927 


4063 


4199 


4335 


4471 


4607 


4743 


4878 


5014 


136 




320 


605150 


505286 


505421 


505557 


505693 


505828 


505964 


50-6099 


506234 


506370 


136 




1 


6505 


6640 


6776 


6911 


7046 


7181 


7316 


7451 


7586 


7721 


135 




2 


7856 


7991 


8126 


8260 


8395 


8530 


8664 


8799 


8934 


9068 


135 




3 


9203 


9337 


9471 


9606 


9740 


9874 


510009 


510143 


510277 


510411 


134 




4 


510545 


510679 


510813 


510947 


511081 


511215 


1349 


1482 


1616 


1750 


134 




5 


1883 


2017 


2151 


2284 


2418 


2551 


2684 


2818 


2951 


3084 


133 




6 


3218 


3351 


3484 


3617 


3750 


3883 


4016 


4149 


4282 


4415 


133 




7 


4548 


4681 


4813 


4946 


5079 


5211 


5344 


5476 


5609 


5741 


133 




8 


5874 


6006 


6139 


6271 


6403 


6535 


6668 


6800 


6932 


7064 132 




9 


7196 


7328 


7460 


7592 


7724 


7855 


7987 


8119 


8251 


8382 132 




330 


518514 


518641 


518777 


518909 


519040 


519171 


519303|ol9434 
520615L520745 


519566)519697 


131 




1 


982S 


9958 


52009C 


520221 


520353 


520484 


520876)521007 


131 




5 


52113$ 


52126S 


1400 


1530 


1661 


1792 


is*22 


2053 


2183 


2314 


131 




a 


2444 


t 257J 


2705 


2835 


2966 


3096 


3226 


335fc 


3486 


.1616 


130 




4 


. 374( 


> 387( 


> 400f 


4136 


4266 


439(j 


452C 


465fc 


4785 


4915 


130 




£ 


504J 


> 5174 


l 5304 


, 6434 


6563 


5693 


6822 


6951 


6081 


6210 


129 




( 


1 633! 


) 646J 


> 659S 


! 6727 


6856 


6985 


7114 


7243 


7372 


7501 


129 




I 


r 763( 


) 7759,' 788* 


i 8016 


8145 


8274 


8405 


8531 


8660 


8788 


12^ 




i 


i 891" 


f 90451 9174 


L 9302 


9430 


9559 


9681 


981E 


9943 


530072 


128 




i 


) 630200 530328|530456 530584 530712 


530840 53096$ 


53109t 


.531223 


1351 


m 




N. 





1 


2 3 4 || 5 


6 


7 


8 


9 





6 



LOGARITHMS 



TT 


1 


2 


3 


4 


6 


6 


7 


8 9 1). 


340 


531479 


531GU7 


531734 


531862 


531990 


532117 


532245 


632372 


532500 532627 128 


l 


2754 


2882 


3009 


3136 


3264 


3391 


3518 


3645 


3772 


3899 


127 


2 


402b 


4153 


4280 


4407 


4534 


4661 


4787 


4914 


5041 


5167 


127 


3 


5294 


5421 


5547 


5674 


6800 


6927 


6053 


6180 


6306 


6432 


126 


4 


6558 


6685 


6811 


6937 


7063 


7189 


7315 


7441 


7567 


7693 


126 


5 


7819 


7945 


8071 


8197 


8322 


8448 


8574 


8699 


8825 


8951 


126 


6 


9076 


9202 


9327 


9452 


9578 


9703 


9829 


9954 


540079 


540204 


125 


7 


540329 


540455 


540580 


540705 


540830 


540955 


541080 541205 


1330 


1454 


125 


8 


1579 


1704 


1829 


1953 


2078 


2203 


2327 2452 


2576 


270] 


125 


9 


2825 


2950 


3074 


3199 


3323 


3447 


3571 3696 


3820 


3944 


124 


350 


544068 


544192 


544316 


544440 


544564 


544688 


544812 


544936 


545060 


5451831: 


1 


5307 


5431 


5555 


5678 


6802 


6925 


6049 


6172 


6296 


6419 


124 


2 


6543 


6666 


6789 


6913 


7036 


7159 


7282 


7405 


7529 


7652 


123 


3 


7775 


7898 


8021 


8144 


8267 


8389 


8512 


8635 


8758 


8881 


123 


4 


9003 


9126 


9249 


9371 


9494 


9616 


9739 


9861 


9984|550106 


123 


5 


550228 


550351 


550473 


550595 


550717 


550840 


550962 


551084 


551206 


1328 


122 


6 


1450 


1572 


1694 


1816 


1938 


2060 


2181 


2303 


2425 


2547 


122 


7 


2668 


2790 


2911 


3033 


3155 


3276 


3398 


3519 


3640 


3762 


121 


8 


3883 


4004 


4126 


4247 


4368 


4489 


4610 


4731 


4852 


4973 


121 


9 


5094 


5215 


5336 


5457 


6578 


6699 


5820 


6940 6061 


6182121 


360 


556303 


556423 


556544 656664 


556785 


556905 


557026 


557146 


557267 


557387 


120 


1 


7507 


7627 


7748 7868 


7988 


8108 


8228 


8349 


8469 


8589 


120 


2 


8709 


8829 


8948 9068 


9188 


9308 


9428 


9548 


9667 


9787 


120 


3 


9907 


060026 


560146560265 


560385 


560504 


560624 


560743 


560863 


560982 


119 


4 


561101 


1221 


1340 


1459 


1578 


1698 


1817 


1936 


2055 2174 


119 


5 


2293 


2412 


2531 


2650 


2769 


2887 


3006 


3125 


3244 


3362 


119 


6 


3481 


3600 


3718 


3837 


3955 


4074 


4192 


4311 


4429 


4548119 


7 


4666 


4784 


4903 


5021 


5139 


5257 


5376 


5494 


5612 


5730118 


8 


5848 


5966 


6084 


6202 


6320 


6437 


6555 


6673 


6791 


6909 118 


9 


7026 


7144 


7262 


7379 


7497 


7614 


7732 


7849 


7967 8084 lib 


370 568202 568319 


568436 


568554 


568671 


568788 


568905 


569023 


5691401569257 


117 


1 


9374 


9491 


9608 


9725 


9842 


9959 


570076 


570193 


570309 570426 


117 


2 


570543 


570660 


570776 


570893 


571010 


571126 


1243 


1359 


1476 


1592 


117 


3 


1709 


1825 


1942 


2058 


2174 


2291 


2407 


2523 


2639 


2755 


116 


4 


2872 


2988 


3104 


3220 


3336 


3452 


3568 


3684 


3800 


3915 


116 


5 


4031 


4147 


4263 


4379 


4494 


4610 


4726 


4841 


4957 


5072 


116 


6 


5188 


6303 


5419 


5534 


6650 


6765 


6880 


6996 


6111 


6226 


115 


7 


6341 


6457 


6572 


6687 


6802 


6917 


7032 


7147 


7262 


7377 


115 


8 


7492 


7607 


7722 


7836 


7951 


8066 


8181 


8295 


8410 


8525 


115 


9 8639 


8754 


8868 


8983 


9097 


9212 


9326 


9441 


9555 


9669 


114 


380 


579784 


579898 


580012 


5801261580241 


580355 


580469 


580583 


580697 


580811 


114 


1 


680925 


581039 


1153 


1267 


1381 


1495 


1608 


1722 


1836 


1950 


114 


2 


2063 2177 


2201 


2404 


2518 


2631 


2745 


2858 


2972 


3085 


114 


3 


3199 


3312 


3 126 


3539 


3652 


3765 


3879 


3992 


4105 


4218 


m 


4 


4331 


4444 


4557 


4670 


4783 


4896 


6009 


' 5122 


5235 


5348 


lit 


5 5461 


5574 


5686 


5799 


5912 


6024 


6137 


6250 


6362 


6475 


113 


6 6587 


6700 


6812 


6925 


7037 


7149 


7262 


7374 


7486 


7599 


112 


7 


7711 


7823 


7935 


8047 


8160 


8272 


8384 


8496 


8608 


8720 


112 


8 


8832 


8944 


9056 


9167 


9279 


9391 


9503 


9615 


9726 


9838 


112 


9 


9950r.«)0061 


590173 


590284 


590396 


590507 


590619 


590730 


590842 590953 112 


390 


591065 


591176 


591287 


591399 


591510 


591621 


591732 


591843 


591955 


592066J111 


1 


2177 


2288 


2399 


2510 


2621 


2732 


2843 


2954 


3064 


3175illl 


2 


3286 


3397 


3508 


3618 


3729 


3840 


3950 


4061 


4171 


4282llll 


3 


4393 


4503 


4614 


4724 


4834 


4945 


6055 


5165 


5276 


6386110 


4 


6496 


5606 


5717 


6827 


5937 


6047 


6157 


6267 


6377 


6487|110 


5 


6597 


6707 


6817 


6927 


7037 


7146 


7256 


7366 


7476 


7586 110 


6 


7695 


7805 


7914 


8024 


8134 


8243 


8353 


8462 


8572 


8681 


110 


7 


8791 


8900 


9009 


9119 


9228 


9337 9446 


9556 


9665 


9774 


109 


8 


9833 


9992 


600101 


600210 


600319 


600428600537 


600646 


600755 600864 


109 


9 600973 


601082 


1191 


1299 


1408 


1517 1625 


1734 


1843| 1951109 


N.| 


1 


2 * 3 


5 6 


7 


8 


B ID. 



OF NUMBERS. 



'N.| o 


'1 


2 


3 | 4 | 


5 


6 


7 8 | 9 | 


D. 




400 (j0201j0 


uU21t)9 


B02277 


5023861602494 


602603 


J02711 


U02819 602928 | 603036 


1U8 




1 


3144 


3253 


3361 


3469 


3577 


3686 


3794 


3902 


4010 


4118 


108 




2 


4226 


4334 


4442 


4550 


4658 


4766 


4874 


4982 


5089 


5197 


108 




3 


6305 


5413 


6521 


6628 


5736 


5844 5951 


6059 


6166 


6274 


108 




4 


6381 


6489 


6596 


6704 


6811 


6919 7026 


7133 


7241 


7348 


107 




5 


7455 


7562 


7669 


7777 


7884 


799l| 8098 


8205 


8312 


8419 


107 




6 


8526 


8633 


8740 


8847 


8954 


9061 9167 


9274 


9381 


9488 


107 




7 


9594 


9701 


9808 


9914 


610021 


610128 610234 


610341 


610447 


610554 


107 




8 


S10660 


610767 


610873 


610979 


1086 


1192 1298 


1405 


1511 


1617 


106 




9 


1723 


1829 


1936 


2042 


2148 


2254 2360 


2466 


2572 


2678 


106 




410 


612784 


612890 


612996 


613102 


613207 


613313 


613419 


613525 


613630 


613736 


100 




1 


3842 


3947 


4053 


4159 


4264 


4370 


4475 


4581 


4686 


4792 


106 




2 


4897 


6003 


6108 


6213 


5319 


5424 


6529 


5634 


5740 


5845 


io. r . 




3 


5950 


6055 


6160 


6265 


6370 


6476 


6581 


6686 


6790 


6895 


105 




4 


7000 


7105 


7210 


7315 


7420 


7525 


7629 


7734 


7839 


7943 


10S 




6 


8048 


8153 


8257 


8362 


8466 


8571 


8676 


8780 


8884 


8989 


105 




6 


9093 


9198 


9302 


9406 


9511 


9615 


9719 


9824 


9928 


620032 


104 




7 


620136 


620240 


620344 


620448 


620552 


620656 


620760 


620864 


620968 


1072 


104 




8 


1176 


1280 


1384 


1488 


1592 


1695 


1799 


1903 


2007 


2110 


104 




9 


2214 


2318 


2421 


2525 


2628 


2732 


2835 


2939 


3042 


3146 


104 




420 


623249 


623353 


623456 


623559 


623663 


623766 


623869 


623973 


624076 


624179 


103 




I 


4282 


4385 


4488 


4591 


4695 


4798 


4901 


6004 


5107 


5210 


103 




2 


5312 


6415 


5518 


6621 


5724 


5827 


5929 


6032 


6135 


6238 


103 




3 


6340 


6443 


6546 


6648 


6751 


6853 


6956 


7058 


7161 


7263 


103 




4 


7366 


7468 


7571 


7673 


7775 


7878 


7980 


8082 


8185 


8287 


102 




5 


8389 


8491 


8593 


8695 


8797 


8900 


9002 


9104 


9206 


9308 


102 




6 


9410 


9512 


9613 


9715 


9817 


9919 


610021 


630123 


630224 


630326 


102 




7 


630428 


630530 


630631 


630733 


630835 


630936 


1038 


1139 


1241 


1342 


102 




8 


1444 


1545 


1647 


1748 


1849 


1951 


2052 


2153 


2255 


2356 


101 




9 


2457 


2559 


2660 


2761 


2862 


2963 


3064 


3165 


3266 


3367 


101 




430 


633468 


633569 


633670 


633771 


633872 


633973 


634074 


634175 


634276 


634376,101 




1 


4477 


4578 


4679 


4779 


4880 


4981 


6081 


5182 


5283 


5383101 




2 


6484 


5584 


5685 


6785 


5886 


5986 


6087 


6187 


6287 


6388 100 




3 


6488 


6588 


6688 


6789 


6889 


6989 


7089 


7189 


7290 


7390100 




4 


7490 


7590 


7690 


7790 


7890 


7990 


8090 


8190 


8290 


8389 


100 




5 


8489 


8589 


8689 


8789 


8888 


8988 


9088 


9188 


9287 


9387 


100 




6 


9486 


9586 


9686 


9785 


9885 


9984 


640084 


640183 640283 


640382 


99 




7 


640481 


640581 


640680 


640779 


640879 


640978 


1077 


11771 1276 


1375 


99 




8 


1474 


1573 


1672 


1771 


1871 


1970 


2069 


2168 2267 


2366 


99 




9 


2465 


2563 


2662 


2761 


2860 


2959 


3058 


3156| 3255 


3354 


99 




440 


643453 


643551 


643650 6- 


643946 644044|644143 


644242 


644340 98 




1 


4439 


4537 


4636 


4734 


4832 


4931 


6029| 5127 


5226 


6324 


98 




2 


6422 


5521 


6619 


6717 


5815 


6913 


6011 


6110 


6208 


6306 


98 




3 


6404 


6502 


6600 


6698 


6796 


6894 


6992 


7089 


7187 


7285 


98 




4 


7383 


7481 


7579 


7676 


7774 


7872 


7969 


8067 


8165 


8262 


98 




5 


8360 


8458 


8555 


8653 


8750 


8848 


8945 


9043 


9140 


9237 


97 




6 


9335 


9432 


9530 


9627 


9724 


9821 


9919 


650016 


650113 


650210 


97 




7 


650308 


650405 


650502 


650593 


650696 


650793 


650890 


0987 


1084 


1181 


97 




8 


1278 


1375 


1472 


1569 


1666 


17621 :859 


1956 


2053 


2150 


97 




9, 2240 2343 


2440 


2536 


2633 


2730] 2826 


2923 


3019 


3116 


97 




450 


653213 


6533-09 


653405 


653502 


653598 


653695 


653791 


653888 


653984 


654080 


96 




1 


4177 


4273 


4369 


4465 


4562 


4658 


4754 


4850 


4946 


6042 


96 




2 


6138 


5235 


5331 


5427 


6523 


6619 


5715 


5810 


5906 


6002 


9G 




a 


6098 


6194 


6290 


6386 


6482 


6577 


6673 


6769 


6864 


6960 


96 




i 


7056 


, 7152 


7247 


7343 


7438 


7534 


7629 


7725 


7820 


7916 


96 




6 


8011 


8107 


8205 


8298 


- 8393 


8488 


8584 


8678 


8774 


887C 


95 




6 


» 896J 


906C 


9155 


925(1 


9346 


9441 


9536 


9631 


9726 


9821 


95 




1 


9911 


> 6G0011 


660106 


660201 


660296 


660391 


660486 


660581 


66067b 


660771 


95 




* 


1 66086E 


» 096C 


( 1055 


115C 


1245 


133S 


1434 


t 152S 


1 1623 


1718 


95 




i 


l 1812 


190' 


2005 


1 2096 2191 


2286 


. 238( 


> 2475 


2569 


2663 


95 




TT 





1 


2 


3 1 4 


1 & 


6 


7 


8 


9 





LOGARITHMS 



N. 


| 1 | 2 


J 


* i 


5 


6 7 8 


9 |D. 


460 


662758(062852 


062947 


603041 


663135 


663230 61 


063512 


003007 


94 


1 


3701 


3795 


3889 


3983 


4078 


4172 


4266| 4360 


4454 


4548 


94 


2 


4642 


4736 


4830 


4924 


6018 


6112 


6206 6299 


6393 


5487 


94 


3 


6581 


6675 


6769 


5862 


6956 


6050 


6145 6237 


6331 


6424 


94 


4 


65 18 


6612 


6705 


6799 


6892 


6986 


7079 


7173 


7266 


7360 


94 


5 


7453 


7546 


7640 


7733 


7826 


7920 


8013 


8106- 8199 


8293 


93 


G 


8386 


8479 


8572 


8665 


8759 


8852 


8945 


90381 9131 


9224 


93 


7 


9317 


9410 


9503 


9596 


9689 


9782 


9875 


9967 


670060 


670153 93 


8 


670246 


670339 


670431 


670524 


670617 


670710 


670802 670895 


0988 


1080 93 


9 

470 


1173 


1265 


1358 


1451 


1543 


1636 


1728| 1821 


1913 


2005 


93 


672098 


672190 


672283 


672375672467 


672560 


672652 672744 


672830 o; 


92 


1 


3021 


3113 


3205 


3297 3390 


3482 


3574 


3666 


3758 


3850 


92 


2 


3942 


4034 


4126 


4218 4310 


4402 


4494 


4586 


4677 


4769 


92 


3 


4861 


4953 


5045 


5137 


6228 


5320 


6412 


5503 


6595 


6687 


92 


4 


5778 


5870 


6962 


6053 


6145 


6236 


6328 


6419 


6511 


6602 


92 


6 


6694 


6785 


6876 


6968 


7059 


7151 


7242 


7333 


7424 


7516 


91 


6 


7607 


7698 


7789 


7881 


7972 


8063 


8154 8245 


8336 


8427 


91 


7 


8518 


8609 


8700 


8791 


8882' 


8973 


9064 9155 


9246 


9337 


91 


8 


9428 


9519 


9610 


9700 


97911 


9882 


9973 '680063 


680154 


680245 


91 


9 


680336 


680426 


680517 


680607 


680698! 


680789 


680879 0970 


1060 


1151 


91 


480 


681241681332 


6814221681513 


681603 


681693 


681784 


681874 


681964 


682055 


90 


1 


2145! 2235 


2326J 2416 


2506 


2596 


2686 


2777 


2867 


2957 


90 


2 


3047 3137 


3227 3317 


3407 


3497 


3587 


3677 


3767 


3857 


90 


3 


3947 


4037 


4127 


4217 


4307 


4396 


4486 4576 


4666 


4756 90 


4 


4845 


4935 


6025 


5114 


6204 


6294 


5383 


6473 


6563 


6652 90 


5 


5742 


6831 


5921 


6010 


6100 


6189 


6279 


6368 


6458 


6547 


89 


6 


6636 


6726 


6815 


6904 


6994 


7083 


7172 


7261 


7351 


7440 


89 


7 


7529 


7618 


7707 


7796 


7886 


7975 


8064J 8153 


8242 


8331 


89 


8 


8420 


8509 


8598 


8687 


8776 


8865 


8953 9042 


9131 


9220 


89 


9 


9309 


9398 


9486 


9575 


9664 


9753 


9841 9930;690019 690107 


89 


490 


690196 


690285 


690373 


690462 


690550 


690639 


690728 


090816 


690905 


690993 


89 


1 


1081 


1170 


1258 


1347 


1435 


1524 


1612 


1700 


1789 


1877 


88 


2 


1965 


2053 


2142 


2230 


2318 


2406 


2494 


2583 


2671 


2759 


88 


3 


2847 


2935 


3023 


3111 


3199 


3287 


3375 


3463 


3551 


3639 


88 


4 


3727 


3815 


3903 


3991 


4078 


4166 


4254 


4342 


4430 


4517 


88 


5 


4605 


4693 


4781 


4868 


4956 


6044 


6131 


5219 


6307 


5394 


88 


6 


6482 


6569 


6657 


6744 


5832 


6919 


6007 


6094 


6182 


6269 


87 


7 


6356 


6444 


6531 


6618 


6706 


6793 


6880 


6968 


7055 


7142 


87 


8 


7229 


7317 


7404 


7491 


7578 


7665 


77521 7839 


7926 


8014 


87 


9 


8101 


8188 


8275 


8362 


8449 


8535 86221 8709 8796 


8883 


87 


500,098970 


699057 


699144 


699231 


699317 


699404 


6994911699578 


099064J699751 


87 


1 9838 


9924 


700011 


700098 


700184 


700271 


700358 


700444 


700531 


700617 


87 


2700704 


700790 


0877 


0963 


1050 


1136 


1222 


1309 


1395 


1482 


86 


31 15C8 


1654 


1741 


1827 


1913 


1999 


2086 


2172 


2258 


2344 


86 


4 243 1 


2517 


2603 


2689 


2775 


2861 


2947 


3033 


3119 


3205 


80 


6 3291 


3377 


3463 


3549 


3635 


3721 


3807 


3893 


3979 


4065 


86 


6 4151 


4236 


4322 


4408 


4494 


4579 


4665 


4751 


48371 4922 


86 


7 


6008 


6094 


5179 


5265 


6350 


5436 


6522 


5607 


6693 6778 


86 


6 


5864 


6949 


6035 


612^1 6206 


6291 


6376 


6462 


65471 6032 


85 


9 


6718 6803 6888 


6974| 7059 


7144 


7229 7315 


7400 


7485 85 


510 


707570 


707655 


707740 


707826 707911 


707996 


708081 


708166 


708251 


708336 


85 


1 


8421 


8506 


8591 


8676 


8761 


8846 


8931 


9015 


9100 


9185 


85 


2 


9270 


9355 


9440 


9524 


9609 


9694 


9779 


9803 


99)3 


710U33 


85 


3 V 710117 


710202 


710287 


710371 


710456 


710540 


710625 


710710 


710794 


0879 


85 


4 


0963i 1048 


1132 


1217 


1301 


1385 


1470 


1654 


1039 


1723 


84 


6 


1807' 1892 1976 


2060 


2144 


2229 


2313 


2397' 2481 


2566 84 


6 


2650 2734 2818 


2902 


2986 


3070 


3154 


3238, 3323| 3407 84 


7 


3491 35751 3659 3742 


3826 


3910 


3994 


4078 4162i 4240 84 


8 


43301 4414. 4497 4581 


4005 


4749 


4833 


4916 


5000, 5084 


84 


9 


61671 5251J 6335| 6418 


6502 


6586 


6069 


6753 


6836; 6920 


84 


N. 


1 1 2 3 


4 


1 & 1 


7 


8 9 |D. 



UF NUMBERS. 



9 



N. 







1 I 



D. 



520 
I 
2 
3 
4 
5 
6 
7 
8 
9 



716003 
6838 
7671 
8502 
9331 

720159 
0986 
1811 
2634 
3456 



716087 
6924 
7754 
8585 
9414 

720242 
1068 
1893 
2716 
3538 



7L6170 
7004 
7837 
8668 
9497 

720325 
1151 
1975 
2798 
3620 



716254 716337 


7088 71711 


7920 8003 


8751 8834 


9580 9663 


720407 720490 


1233 


1316 


2058 


2140 


2881 


2963 


3702 


3784 



716421 
7254 
8086 
8917 
9745 

720573 
1398 
2222 
3045 
3866 



7338 
8169 
9000 
9828 
720655 
1481 
2305 
3127 
3948 



530,724276 



5095 
5912 
6727 
7541 
8354 
9165 
9974 

8 730782 

9 1589 



540 
1 

2 
3 
4 
5 
6 
7 
8 
9 



724358 
5176 
5993 1 
6809 
7623 
8435 
9246 

730055 
0863 
1669 



724440 
5258 
6075 
6890 
7704! 
85161 
9327 

730136 
0944 
1750 



724522 
5340 
6156 
6972 
7785 
8597 
9408 

730217 
1024 
1830 



724601 
5422 
6238 
7053 
7866 
8678 
9489 

730298 
1105 
1911 



732394 
3197 
3999 
4800 
6599 
6397 
7193 
7987 
8781 
9572 



732474 
3278 
4079 
4880 
6679 
6476 
7272 
8067 
8860 
9651 



732555 
3358 
4160 
4960 
6759 
6556 
7352 
8146 
8939 
9731 



732635 
3438 
4240 
5040 
5838 
6635 
7431 
8225 
9018 
9810 



732715 
3518 
4320 
5120 
6918 
6715 
7511 
8305 
9097 
9889 



550 


740363 


1 


1152 


1 


1939 


3 


2725 


4 


3510 


5 


4293 


6 


5075; 


7 


58551 


8 


6034 ! 


9 


7412 



(40442 
1230 
2018 
2804 
3588 
4371 
5153 
5933 
6712 
7489 



560 
1 



748183 
8963 
9736 
3 750508 
1279 
2048 
2816, 
3583 
4348i 
5112 



j70 
1 
2 
3 
4 
5 
6 
7 
8 
9 

tr. 



755875 
6636 
7390, 
8155 
8912 
9668 

760422 
1176 
1928 
2679 




748266 
9040 
9814 

750580 
1356 
2125 
2893 
3660 
4425 
5189 

755951 
6712 
7472 
8230 
8988 
9743 

760498 
1251 
2003 
2754 



740521 
1309 
2096 
2882 
3667 
4449 
5231 
6011 
6790 
7567 



740600 
1388 
2175 
2961 
3745 
4528 
5309 
6089 
6808 
7645 



,740678 
1467 
2254 
3039 
3823 
4606 
5387 
6167 
6945 
7722 



748343 
9118 
9891 

750663 
1433 
2202 
2970 
3736 
4501 
5265 



748421 
9195 
9968 

750740 
1510 
2279 
3047 
3813 
4578 
5341 



756027 
6788 
7548 
8306 
9063 
9819 

760573 
1326 
2078 
2829 



756103 
6864 
7624 
8382 
9139 
9894 

760649 
14)2 
2153 
2904 



748498 
9272 

750045 
0817 
1587 
2356 
3123 
3889 
4654 
5417 

756r8"0 
6940 
7700 
8458 
9214 
9970 

760724 
1477 
2228 
2978 



1 I 2 



724685 
6503 
6320 
7134 
7948 
8759 
9570 



8336 
9165 



7165815 
7421 
8253 
9083 
9911 

720738 720821 
1563 1646 
2387! 2469 
3209 3291 
4030 4112 



16671 716754 83 
75041 7587 83 



8419; 

9248 



9994|720077! 83 



0903 
1728 
2552 
3374 
4194 



724767 
6585 
6401 
7216 
8029 
8841 
9651 



724849 
5667 
6483 
7297; 

8110: 

8922 1 
9732 



724931 
5748 
6564 
7379 
8191 
9003 
9813 



730378|730459!73054O 730621 
1186 1266| 1347; 1428 
1991 2072 2152, 2233 



732796 
3598 
4400 
5200 
6998 
6795 
7590 
8384 
9177 
9968 



732876 
3679 
4480 
6279 
6078 
6874 
7670 
8463 
9256 

740047 



732956 


733037 


3759 


3839 


4560 


4640 


6359 


5439 


6157 


6237 


6954 


7034 


7749 


7829 


8543 


8022 


9335 


9414 


740126 


740205 



725013 
5830 
6646 
7460 
8273 
9084 
9893 

730702 
1508 
2313 

733117 
3919 
4720 
5519 
6317 
7113 
7908 
8701 
9493 

740284 



82 
82 
82 
81 
81 
81 
81 
81 
81 
81 
80 
80 
80 
80 
80 
80 
79 
79 
79 
79 



740757 740836 



1546 

2332 
3118 
3902 
4684 
5465 
6245 
7023 
7800 



1624 
2411 
3196 
3980 
4762 
5543 
6323 
7101 
7878 



748576 
9350 

750123 
0894 
1664 
2433 
3200 
3966 
4730 
6494 



748653 
9427 

750200 
0971 
1741 
2509 
3277 
4042 
4807, 
6570 



740915 


740994 


1703 


1782 


2489 


2568 


3275 


3353 


4058 


4136 


4840 


4919 


5621 


5699 


6401 


6479 


7179 


7256 


7955 


8033 



1860 
2647 
3431 
4215 
4997 
5777 
6556 
7334 
8110 



748731 748808 
9504 9582 

750277 750354 
1048 1125 
1818 1895 
2586 1 2663 
3353 3430 
4119' 4195 
4883] 4960 
5646 6722 



748885 
9659 

750131 
1202 
1972 
2740 
3506 
4272 
5036 
5799 



79 
79 
79 
78 
78 
78 
78 
78 
78 
78 
77 
77 
77 
77 
77 
77 
77 
77 
76 
76 



75625S 
7016 
7775 
8533 
9290 



756332 
7092 
7851 
8609 
9366 



700045:760121 



0799 
1552 
2303 
3053 



0875 
1627 
2378 
3128 
6 



7£ 6408 756484 
7168, 7244 
7927i 8003 
8685 8761 
9441 1 9517 

760196 760272 760347; 
0950 1025 1101 
1702, 1778 1853 
2453: 2529 2604 
3203 3278 3353 



7565601 76 
7320 76 
8079 76 
8836 1 76 
9592i 76 
75 



75 
75 
7S 
76 



9 ID, 



11 



10 










LOGARITHM^ 










~' U 2 3 5 7 | 8 | 9 .,!>■ 


580j70342)- 


703503 


703578 7o3o53 703727 


|763802 703877 Ib'Sjyi 


704027 7 


I 


417t 


4251 


43 21 


4400 4475 


455C 


4624 4699 


4774 


4848 


lb 


5 


S 4923 


4998 


5072 


514TJ 5221 


5291 


6370| 5445 


I 5520 


6594 


75 


a 


6669 


5743 


68 IS 


5892 


6966 


6041 


6115 


6190 


6264 


6338 


74 


4 


6413 


6487 


6502 


603 1 


6710 


6785 


685S 


6933 


7007 


7082 


74 


5 


7156 


7230 


7304 


737S 


7453 


7527 


7601 


7675 


774U 


7823 


74 


fa 


7898 


7972 


8O40 


812C 


8194 


8268 


8342 


8416 


8490 


8564 


74 


7 


8638 


8712 


8780 


8860 


8934 


9008 


9082 


9156 


9230 


9303 


74 


8 


9377 


9451 


9525 


9599 


9673 


9740 


9820 


9894 


9968 


770042 


74 


9 


770115 


770189 


770263 


770336 


770410 


770484 


770557 


770031 '770705 


0778 


74 


590 


770852 


770926 


770999 


771073 


771146 


771220 


7712931771367 


771440 771514 


74 


1 


1587 


1661 


1734 


1808 


1881 


1955 


2028 2102 


2175 


2248 


73 


2 


2322 


2395 


2468 


2542 


2615 


2688 


2762! 2835 


2908 


2981 


73 


3 


3055 


3128 


3201 


3274 


3348 


3421 


3494 3567 


3040 


3713 


73 


4 


3786 


3860 


3933 


4006; 4079 


4152 


4225 4298 


4371 


4444 


73 


5 


4517 


4590 


4663 


4736 


4809 


4882 


49551 5028 


5100 


6173 


73 


6 


6246 


6319 


5392 


5405 


5538 


5610 


5083 ( 5756 


6829 


5902 


73 


7 


5974 


6047 


6120 


6193 


6265 


6338 


6411 6483 


6556 


6629 


73 


8 


6701 


6774 


6846 


6919 


6992 


7004 


7137 7209 


7282 


7354 


73 


9 


7427 


7499 


7572| 7644 


7717 


7789, 7802 


7934 


8006 


8079 


72 


600 


778151 


778224 


778296 


778368 


778441 


778513 


778585 


778658 


778730778802 


72 


1 


8874 


8947 


9019 


9091 


9163 


9236 


930S 


9380 


94521 9524 


72 


2 


9596 


9669 9741 


9813 


9885 


9957 


780029 


780101 


780173 780245 


72 


3 


780317 


780389780461 


780533 


780005 


780677 


0749 


0821 


0893 


0965 


72 


4 


1037 


1109 


1181 


1253 


1324 


1396 


14458 


1540 


1612 


1684 


72 


5 


1755 


1827 


1899 


1971 


2042 


2114 


2186 


2258 


2329 


2401 


72 


6 


2473 


2544 


2616 


2688 


2759 


2831 


2902 


2974 


3046 


3117 


72 


7 


3189 


3260 


3332 


3403 


3475 


3546 


3618 


3689 


3761 


3832 


71 


8 


3904 


397-5 


4046 


4118 


4189 


4261 


4332 


4403 


4475 


4546 


71 


9 


4617 


4689 


4760 


4831 


4902 


4974 


6045 


5116 


5187 


5259 


71 


610 


785330 785401 


785472 


785543 


785615 


785086 


785757 78582 


71 


1 


6041 6112 


6183 


6254 


6325 


6396 


6467 


6538 0609 


6680 


71 


2 


6751) 6822 


6893 


6964 


7035 


7106 


7177 


7248 


7319 


7390 


71 


3 


7460 7531 


7602 


7673 


7744 


7815 


7885 


7956 


8027 


8098 


71 


4 


8168! 8239 


8310 


8381 


8451 


8522 


8593 


8663! 8734 


8804 


71 


5 


8875 8946 


9016 


9087 


9157 


9228 


9299 


9369 9440 


9510 


71 


6 


9581 


9651 


9722 


9792 


9863 


9933 


790004 


790074 


790144 


790215 


70 


7 


790285 


790356 


790426 


790496 


790567 


790637 


0707 


0778 


0848 


0918 


70 


8 


0988 


1059 


1129 


1199 


1269 


1340 


1410 


1480 


1550 


1620 


70 


9 


1691 


1761 


1831 


1901 


1971 


2041 


2111 


2181 


2252 


2322 


70 


620 


792392 


792462 


792532 


792602 


792672 


792742 


792812 7! 


792952 


793022 


70 


1 


3092 


3162 


3231 


3301 


3371 


3441 


3511 


3581 


3651 


3721 


70 


2 


3790 


3860 


3930 


4000 


4070 


4139 


4209 


4279 


4349 


4418 


70 


3 


4488 


4558 


4627 


4697 


4767 


4836 


4906 


4976 


5045 


5115 


70 


4 


5185 


6254 


5324 


6393 


6463 


5532 


5602 


5672 


5741 


6811 


70 


6 


5880 


6949 


6019 


6088 


6158 


6227 


6297 


6366 


6436 


6505 


69 


6 


6574 


6644 


6713 


6782 


6852 


6921 


6990 


7060 


7129! 7198 


69 


7 


7268 


7337 


7406 


7475 


7545 


7614 


7683 


7752 


7821 7890 


69 


8 


7960 


8029 


8098 


8167 


8236 


8305 


8374 


8443 


8513 8582 


09 


9 


8651 


8720 


8789 


8858 


8927 


8996 


9065 9134| 9203! 9272 


69 


630 


799341 


799409 


799478 


799547 


799616 


799685 


799754 


799823 


799892 


799961| 69 


1 


800029 


800098 


800167 


800236 


800305 


800373 


800442 


800511 


800580 


800648 


69 


2 


0717 


078-6 


0854 


0923 


0992 


1061 


1129 


1198 


1266 


1335 


69 


3 


1404 


1472 


1641 


1009 


1678 


1747 


1816 


1884 


1952 


2021 


69 


4 


2089 


2158 


2226 


2295 


2363 


2432 


2500 


2568 


2637 


2705 


68 


5 


2774 


2842 


2910 


2979 


3047 


3116 


3184 


3252 


3321 


3389 


68 


6 


3457 


3525 


3594 


3662 


3730 


3798 


3867 


3935 4003 


4071 


68 


7 


4139 


4208 


4276 


4344 


4412 


4480 


4548 


46161 1685 


4753 


68 


8 


4821 


4889 


4967 


5025 


5093 


5161 


6229 


5297 5365 


5433 


68 


9 


6501 


5569 


6637 


6705 


6773 


5841 


5908 


5976 6044 


6112 


68 


N.| 





1 


2 


3 


4 II 


6 | 7 


9 \D. 



OF NUMBERS. 



11 



N. 





1 


2 


3 


4 


6 


6 


7 8 9 | D. 


640 


806180 


806248 


806316 


806384 


806451 


8065 li> 


806587 


806655 806723 806790 68 


1 


6858 


6926 


6994 


7061 


7129 


7197 


7264 


7332 


7400 


7467 


68 


2 


7535 


7603 


7670 


7738 


7806 


7873 


7941 


8008 


8076 


8143 


68 


3 


8211 


8279 


8346 


8414 


8481 


8549 


8616 


8684 


8751 


8818 


67 


4 


8886 


8953 


9021 


9088 


9156 


9223 


9290 


9358 


0425 


9492 


67 


5 


9560 


9627 


9694 


9762 


9829 


9896 


9964 


810031 


810098 


810165 


67 


6 


810233 


810300 


810367 


810434 


810501 


810569 


810636 


0703 


0770 


0837 


67 


7 


0904 


0971 


1039 


1106 


1173 


1240 


1307 


1374 


1441 


1608 67 


8 


1575 


1642 


1709 


1776 


1843 


1910 


1977 


2044 


2111 


2178 67 


9 


2245 


2312 


2379 


2445 


2512 


2579 


2646 


2713 


2780 


2847| 67 


(150 


812913 


812980 


813047 


813114 


813181 


813247 


813314 


813381 


813448 8] 


67 


1 


3581 


3648 


3714 


3781 


3848 


3914 


3981 


4048 


4114 


4181 


67 


2 


4248 


4314 


4381 


4447 


4514 


4581 


4647 


4714 


4780 


4847 


67 


3 


4913 


4980 


5046 


5113 


5179 


5246 


5312 


6378 


6445 


6511 


66 


4 


6678 


5644 


6711 


5777 


5843 


5910 


6976 


6042 


6109 


6175 


66 


6 


6241 


6308 


6374 


6440 


6506 


6573 


6639 


6705 


6771 


6838 


66 


6 


6904 


6970 


7036 


7102 


7169 


7235 


7301 


7367 


7433 


7499 


66 


7 


7565 


7631 


7698 


7764 


7830 


7896 


7962 


8028 


8094 


8160 


66 


8 


8226 


8292 


8358 


8424 


8490 


8556 


8622 


8688 


8754 


8820 


66 


9 


8885 


8951 


9017 


9083 


9149 


9215 


9281 


9346 


9412 


9478 


66 


660 


819544181 


819676 


819741 


819807 


819873 


819939 


820004 


820070 


820136 


66 


1 


820201 


820267 


820333 


820399 


820464 


820530 


820595 


0661 


0727 


0792 


66 


2 


0858 


0924 


0989 


1055 


1120 


1186 


1251 


1317 


1382 


1448 


66 


3 


1514 


1579 


1645 


1710 


1775 


1841 


1906 


1972 


2037 


2103 


65 


4 


2168 


2233 


2299 


2364 


2430 


2495 


2560 


2626 


2691 


2756 


65 


5 


2822 


2887 


2952 


3018 


3083 


3148 


3213 


3279 


3344 


3409 


66 


6 


3474 


3539 


3605 


3670 


3735 


3800 


3865 


3930 


3996 


4061 


65 


7 


4126 


4191 


4256 


4321 


4386 


4451 


4616 


4581 


4646 


4711 


65 


8 


4776 


4841 


4906 


4971 


5036 


5101 


5166 


5231 


5296 


5361 


65 


9 


6426 


5491 


5556 


5621 


5686 


5751 


5815 


6880 


6945 


6010 


65 


670 


826075 


826140 8: 


826269 


826334 


826399 


826464 


826528 


826593 


826658 


65 


1 


6723 


6787 


6852 


6917 


6981 


7046 


7111 


7175 


7240 


7305 


66 


2 


7369 


7434 


7499 


7563 


7628 


7692 


7767 


7821 


7886 


7951 


65 


3 


8015 


8080 


8144 


8209 


8273 


8338 


8402 


8467 


8531 


8595 


64 


4 


8660 


8724 


8789 


8853 


8913 


8982 


9046 


9111 


9175 


9239 


64 


5 


9304 


9368 


9432 


9497 


9561 


9625 


9690 


9754 


9818 


9882 


64 


6 


9947 


830011 


830075 


830139 


830204 


830268 


830332 


830396 


830460 


830525 


64 


7 


830589 


0653 


0717 


0781 


0845 


0909 


0973 


1037 


1102 


1166 


64 


8 


1230 


1294 


1358 


1422 


1486 


1550 


1614 


1678 


1742 


1806 


64 


9 


1870 


1934 


1998 


2062 


2126 


2189 


2253 


2317 


2381 


2446 64 


680 832509 


832573 


832637 


832700 83 


832828 


832892 


832956 


833020 8! 


64 


1 


3147 


3211 


3275 


3338 


3402 


3466 


3530 


3593 


3657 


3721 


64 


2 


3784 


3848 


3912 


3975 


4039 


4103 


4166 


4230 


4294 


4357 


64 


3 


4421 


4484 


4548 


4611 


4675 


4739 


4802 


4866 


4929 


4993 


64 


4 


5056 


5120 


6183 


5247 


5310 


6373 


6437 


6500 


6564 


5627 


63 


5 


5691 


5754 


6817 


5881 


6944 


6007 


6071 


6134 


6197 


6261 


63 


6 


6324 


6387 


6451 


6514 


6577 


6641 


6704 


6767 


6830 


6894 


63 


7 


6957 


7020 


7083 


7146 


7210 


7273 


7336 


7399 


7462 


7525 


63 


8 


7588 


7652 


7715 


7778 


7841 


7904 


7967 


8030 


8093 


8156 


63 


9 

690 


8219 


8282 


8345 


8408 


8471 


8634 


8597 


8660 


8723 


8786 


63 


838849 


838912 


838975 


839038 


839101 


839164 


839227 


839289 


839352 


8394151 6 


1 


9478 


9541 


9604 


9667 


9729 


9792 


9856 


9918 


9981 


840043 


63 


2 


840106 


840169 


840232 


840294 


840357 


840420 


840482 


840545 


840608 


0671 


63 


3 


0733 


0796 


0859 


0921 


0984 


1046 


1109 


1172 


1234 


1297 


63 


4 


1359 


1422 


1485 


1547 


1610 


1672 


1735 


1797 


1860 


1922 


63 


6 


1985 


2047 


2110 


2172 


2235 


2297 


2360 


2422 


2484 


2547 


62 


6 


2609 


2672 


2734 


2796 


2859 


2921 


2983 


3046 


3108 


3170 


62 


7 


3233 


3295 


3357 


3420 


3482 


3544 


3606 


3669 3731 3793 


62 


8 


3855 


3918 


3980 


4042 


4104 


4166 


4229 


4291 43531 4415 


62 


9 


4477 


4539 


4601 


4664 


4726 


4788 


4850 


4912! 4974! 6036 


62 


TT~ 


1 1 2 


3 | 4 


1 & 1 


7 


8 9 


D 



12 










LOGARITHMS 












N. | 


3 1 


5 6 


7 


8 


9 | D. 


700 8450U8j845 1 0O|845222 


845284 845146 


845408 


845470 


345532 


845594 8, 


02 


1 


5718' 6780 


5842 


6904 


5966 


6028 


6090 


6151 


6213 


6275 


62 


2 


6337 


6399 


6461 


6523 


6585 


6646 


6708 


6770 


6832 


6894 


62 


3 


6955 


7017 


7079 


7141 


7202 


7264 


7326 


7388 


7449 


7511 


62 


4 


7573 


7634 


7696 


7758 


7819 


7881 


7943 


8004 


8000! 8128 


62 


5 


8189 


8251 


8312 


8374 


8435 


8497 


8559 


8620 


8682 


8743 


62 


6 


8805 


8806 


8928 


8989 


9051 


9112 


9174 


9235 


9297 


9358 


61 


7 


9419 


9481 


9542 


9604 


9665 


9726 


9788 


9849 


9911 


9972 


61 


8 


850033 


850O'J5 850150 


850217 


850279 


850340 


850401 


850462 


850524 


850585 


61 


9 


0646 


0707| 0769 


0830 


0891 


0952 


1014 


1076 


1136 


1197 


61 


710 


851258 


851320 


851381 


851442 


851503 


851564 


851625 


85108G 


851747 


851809 


61 


1 


1870 


1931 


1992 


2053 


2114 


2175 


2236 


2297 


2358 


2419 


61 


2 


2480 


2541 


2602 


2663 


27M 


2785 


2846 


2907 


2908 


3029 


61 


3 


3090 


3150 


3211 


3272 


3333 


3394 


3455 


3510 


3577 


3637 


61 


4 


3698 


3759 


3820 


3881 


3941 


4002 


4003 


4124 


4185 


4245 


61 


6 


4306 


4367 


4428 


4488 


4549 


4610 


4670 


4731 


4792 


4852 


61 


6 


4913 


4974 


5034 


6095 


5156 


5216 


5277 


5337 


6398 


5459 


61 


7 


5619 


5580 


6640 


5701 


5761 


5822 


5882 


6943 


6003 


6064 


61 


8 


6124 


6185 


6245 


6306 


6366 


6427 


6487 


6548 


6608 


6668 


60 


» 


6729 


6789 


6850 


6910 


6970 


7031 


7091 


7152 


7212 


7272 


60 


720 


857332 857: 


867453 


857613'8! 


857634 


857694 


857755 85 


60 


1 


7935 


7995 


8056 


8116 


8176 


8236 


8297 


8357 


8417 


8477 


60 


2 


8537 


8597 


8657 


8718 


8778 


8838 


8898 


8958 


9018 


9078 


60 


3 


9138 


9198 


9258 


9318 


9379 


0439 


9499 


9559 9619 


9679 


60 


4 


9739 


9799 


9859 


9918 


9978 


860038 


860098 


860158860218 


860278 


60 


6 


860338 


860398 


860458 


860518 


860578 


0637 


0697 


0757 0817 


0877 


60 


6 


0937 


0996 


1056 


1116 


1176 


1236 


1295 


1355 1415 


1475 


60 


7 


1634 


1594 


1654 


1714 


1773 


1833 


1893 


1952 2012 


2072 


60 


8 


2131 


2191 


2251 


2310 


2370 


2430 


2489 


2549 2008 


2668 


60 


9 


2728 


2787 


2847 


2906 


2966 


3025 


3085 


3144 


3204 


3203 


60 


730 


863323 


863382 


863442 


863501 


863561 


863 020 


863680 


863739 


863799 8t 


59 


1 


3917 


3977 


4036 


4096 


4155 


4214 


4274 


4333 


4392 


4452 


69 


2 


4511 


4670 


4630 


4689 


4748 


4808 


4867 


4926 


4985 


6045 


59 


3 


5104 


5163 


6222 


5282 


5341 


6400 


6459 


6519 


5578 


6637 


69 


4 


5696 


6755 


5814 


6874 


5933 


6992 


6051 


6110 


6169 


6228 


59 


5 


6287 


6346 


6406 


6465 


6524 


6583 


6642 


6701 


6760 


6819 


69 


6 


6878 


6937 


6996 


7055 


7114 


7173 


7232 


7291 


7350 


7409 


59 


7 


7467 


7626 


7585 


7644 


7703 


7762 


7821 


7880 


7939 


7998 


59 


8 


8056 


8116 


8174 


8233 


8292 


8350 


8409 


8408 


8527 


8586 


69 


9 


8644 


8703 


8762 


8821 


8879 


8938 


8997 


9050 


9114 


9173 


59 


740 


869232 


869290 


869349 


869408 8< 


869525 


869584 


809042 


869701 


869760 


59 


1 


9818 


9877 


9935 


9994 


870053 


870111 


870170 


870228 


870287 


870345 


59 


2 


870404 


870462 


870521 


870579 


0638 


0696 


0755 


0813 


0872 


0930 


58 


3 


0989 


1047 


1106 


1164 


1223 


1281 


1339 


1398 


1456 


1515 


58 


4 


1573 


1631 


1690 


1748 


1806 


1865 


1923 


1981 


2040 


2098 


58 


6 


2156 


2215 


2273 


2331 


2389 


2448 


2506 


2564 


2622 


2681 


68 


6 


2739 


2797 


2865 


2913 


2972 


3030 


3088 


3146 


3204 


3262 


58 


7 


3321 


3379 


3437 


3495 


3553 


3611 


3609 


3727 


3785 


3844 


58 


8 3902 


3960 


4018 


4076 


4134 


4192 


4250 


4308 


4300 


4424 


68 


9| 4482 


4540 


4598 


4656 


4714 


4772 


4830 


4888 


4945 


5003 


58 


750 


875061 


875119 


875177 


875235 


875293 


876351 


875409 


875406 


875524 


8755821 5 


1 


5640 


6698 


6766 


6813 


6871 


6929 


6987 


6045 


6102 


6160 


58 


2 


6218 


6276 


6333 


6391 


6449 


6507 


6564 


6622 


6680 


6737 


58 


3 


6795 


6853 


6910 


6968 


7026 


7083 


7141 


7199 


7256 


7314 


58 


4 


7371 


7429 


7487 


7644 


7602 


7659 


7717 


7774 


7832 


7889 


58 


6 


7947 


8004 


8062 


8119 


8177 


8234 


8292 


8349 


8407 


8464 


57 


6 


8522 


8579 


8637 


8694 


8762 


8809 


8866 


8924 


8981 


9039 


57 


7 


9096 


9153 


9211 


9268 


9325 


9383 9440 


9497 


9555 


9612 


57 


8 


9669 


9726 


9784 


9841 


9898 


9956'SS0013 


8800701880127 


880185 


57 


9 


880242 


8*0299' 880356 


8804.131880471 


880528 0586 


06421 0699 


0750 


57 


te 





1 | 2 


3 4 


6 6 


I 7 8 


i> 



OP NUMBERS. 



13 



N. 





X 


2 


3 


4 1 


5 


6 


7 8 


9 | D. 


765 


880814 


880871 


880928 


880985 


881042 


881099 


881156 


881213 881271 


8813281 67 


1 


1385 


1442 


1499 


1556 


1613 


1670 


1727 


1784 


1841 


1898 


67 


2 


1955 


2012 


2069 


2126 


2183 


2240 


2297 


2354 


2411 


2468 


67 


3 


2525 


2581 


2638 


2695 


2752 


2809 


2866 


2923 


2980 


3037 


67 


4 


3093 


3150 


3207 


3264 


3321 


3377 


3434 


3491 


3548 


3605 


57 


5 


3661 


3718 


3775 


3832 


3888 


3945 


4002 


4059 


4115 


4172 


67 


6 


4229 


4285 


4342 


4399 


4455 


4512 


4569 


4625 


4682 


4739 


57 


7 


4795 


485'2 


4909 


4965 


6022 


5078 


6135 


5192 


5248 


6305 


57 


8 


5361 


6418 


5474 


6531 


5587 


6644 


6700 


6757 


5813 


6870 


57 


9 


6926 


0383 


6039 


6096 


6152 


6209 


6265 


6321 


6378 


6434 


66 


770 


886491 


88054»7 


88<A504 


886660 


886716 


886773 


880829.88 


886942 


«86fi<9S 


66 


1 


7054 


7111 


7167 


7223 


7280. 


7336 


7392 


744i> 


7505 


7561 


56 


2 


7617 


7674 


7730 


7786 


7842. 


7898 


7955 


8011 


8067 


8123 


56 


3 


8179 


8236 


8292 


8348 


8404 


8460 


8516 


8573 


8629 


8685 


56 


4 


8741 


8797 


8853 


8909 


8965 


9021 


9077 


9134 


9190 


9246 


56 


5 


9302 


9358 


9414 


9470 


9526 


9582 


9638 


9691 


9750 


9806 


56 


6 


9862 


9918 


9974 


890030 


890086 


890141- 


890197 


890253 


890309 


890365 


56 


7 


890421 


890477 


890533 


0589 


0645 


0700 


0756 


0812 


0808 


0924 


56 


8 


0980 


1035 


1091 


1147 


1203 


1259 


1314 


1370 


1426 


1482 


66 


9 


1537 


1593 


1649 


1705 


1760 


1816 


1872 


1928 


1983 


2039 


56 


780 


892095 


892150 


892206 8i 


892317 


892373 8! 


892484 


892540 8i 


56 


1 


2651 


2707 


2762 


2818 


2873 


2929 


2985 


3040 


3096 


3151 


56 


2 


3207 


3202 


3318 


3373 


3429 


3484 


3540 


3595 


3651 


3706 


56 


3 


3762 


3817 


3873 


3928 


3984 


4039 


4094 


4150 


4205 


4261 


55 


4 


4316 


4371 


4427 


4482 


4538 


4693 


4648 


4704 


4759 


4814 


55 


6 


4870 


4925 


4980 


5036 


6091 


5146 


6201 


6257 


5312 


6367 


66 


6 


54'23 


5478 


6533 


5588 


5644 


6699 


6754 


6809 


6864 


6920 


65 


7 


5975 


6030 


6085 


6140 


6195 


6251 


6306 


6361 


6416 


6471 


55 


8 


6526 


6581 


6636 


6692 


6747 


6802 


6857 


6912 


6967 


7022 


55 


9 


7077 


7132 


7187 


7242 


7297 


7352 


7407 


7462 


7517 


7572 


55 


790 


897627 


897682 


897737 


897792 


897847 


897902 


897959 


898012 


898067 


898122 


55 


1 


8176 


8231 


8286 


8341 


8396 


8451 


8506 


8561 


8615 


8670 


55 


2 


8725 


8780 


8835 


8890 


8944 


8999 


9054 


9109 


9164 


9218 


55 


3 


9273 


9328 


9383 


9437 


9492 


9547 


9602 


9656 


9711 


9766 


55 


4 


9821 


9675 


9930 


9985 


900039 


900094 


900149 


900203 


900258 


900312 


56 


5 


900367 


900422 


900476 


900531 


0586 


0640 


0695 


0749 


0804 


0859 


55 


6 


0913 


0968 


1022 


1077 


1131 


1186 


1240 


1295 


1349 


1404 


55 


7 


1458 


1513 


1567 


1622 


1676 


1731 


1785 


1840 


1894 


1948 


54 


8 


2003 


2057 


2112 


2166 


2221 


2275 


2329 


2384 


2438 


2492 


54 


9 2547 


2601 


2655 


2710 


2764 


2818 


2873 


2927 


2981 


3036 


54 

54 


300 


903090 


903144 


903199 


903253 


903307 


903361 


903416 


903470 


903524 


903678 


1 


3633 


3687 


3741 


3795 


3849 


3904 


3958 


4012 


4066 


4120 


54 


2 


4174 


4229 


4283 


4337 


4391 


4445 


4499 


4553 


4607 


4661 


54 


3 


4716 


4770 


4824 


4878 


4932 


4986 


6040 


5094 


6148 


6202 


54 


4 


5256 


5310 


5364 


6418 


5472 


5526 


6580 


6634 


5688 


6742 


54 


5 


67% 


6850 


5904 


5958 


6012 


6066 


6119 


6173 


6227 


6281 


54 


6 


6335 


6389 


6443 


6497 


6551 


6604 


6658 


6712 


6766 


6820 


54 


7 


6874 


6927 


6981 


7035 


7089 


7143 


7196 


7250 


7304 


7358 


54 


8 


7411 


7465 


7519 


7573 


7626 


7680 


7734 


7787 


7841 


7895 


54 


9 


7949 


8002 


6056 


8110 


8163 


8217 


8270 


8324 


8378 


8431 54 


810 


908485 


908539 


908592 


908646 


908699 


908753 


908807 


908860 


908914 9( 


54 


1 


9021 


9074 


9128 


9181 


9235 


9289 


9342 


9396 


9449 


9503 


54 


2 


9556 


9610 


9663 


9716 


9770 


9823 


9877 


9930 


9984 


910037 


63 


3 


310091 


910144 


910197 


910251 


910304 


910358 


910411 


910464 


910518 


0571 


63 


4 


0624 


0678 0731 


0784 


0838 


0891 


0944 


0998 


1061 


1104 


53 


5 


1158 


1211 


1264 


1317 


1371 


1424 


1477 


1530 


1584 


1637 


63 


6 


1690 


1743 


1797 


1850 


1903 


1956 


2009 


2063 


2116 


2169 


53 


7 


2222 


2275 


2328 


2381 


2435 


2488 


2541 


2594 


2647 


2700 


53 


8 


2753 


2806 


28591 2913 


2966 


3019 


3072 


3125 


3178 


3231.53 


9 


3284 


3337 


33901 3443 


3496 


3549 


3602 


3655 


3708 


3761| 53 


"NT 


o 


1 


4 


| 5 | 6 


1 


8 


9 , 


£J 



ir 



14 










LOGARITHMS 












N.| 


1 1 


3 


4 || 6 


7 


8 


9 


IV 


820 1)13814 


913807 


913920 


913973 


014026 


914079 


91-1132 


914184 


914237 


914290 


53 


1 


4343 


4396 


4449 


4502 


4555 


4608 


4660 


4713 


4766 


4819 


53 


2 


4872 


4925 


4977 


6030 


6083 


6136 


6189 


6241 


5294 


6347 


53 


3 


6400 


5453 


5505 


5558 


5611 


6664 


5716 


6769 


5822 


6875 


53 


4 


6927 


6980 


6033 


6085 


6138 


6191 


6243 


6296 


6349 


6401 


53 


6 


6454 


6507 


6559 


6612 


6664 


6717 


6770 


6822 


6875 


6927 


63 


G 


6980 


7033 


7085 


7138 


7190 


7243 


7295 


7348 


7400 


7453 


53 


7 


7506 


7558 


7611 


7663 


7716 


7768 


7820 


7873 


7925 


7978 


62 


8 


8030 


8083 


8135 


8188 


8240 


8293 


8345 


8397 


8450 


8502 


52 


9 


8555 


8607 


8659 


8712 


8764 


8816 


.8869 


8921 8973 


9026 


52 


830 


919078 


919130 


919183 


919235 


919287 


919340 


919392 


919444 


919496 


919549 


62 


1 


9601 


9653 


9706 


9758 


9810 


9862 


9914 


9967 


920019 


920071 


52 


2 


920123 


920176 


920228 


920280 


920332 


920384 


920436 


920489 


0541 


0593 


62 


3 


0645 


0697 


0749 


0801 


0853 


0906 


0958 


1010 


1062 


1114 


52 


4 


1166 


1218 


1270 


1322 


1374 


1426 


1478 


1530 


1582 


1634 


52 


6 


1686 


1738 


1790 


1842 


1894 


1946 


1998 


2050 


2102 


2154 


52 


6 


2206 


2258 


2310 


2362 


2414 


2466 


2518 


2570 


2622 


2674 


62 


7 


2725 


2777 


2829 


2881 


2933 


2985 


3037 


3089 


3140 


3192 


52 


8 


3244 


3296 


3348 


3399 


3451 


3503 


3555 


3607 


3658 


3710 


52 


9 


3762 


3814 


3865 


3917 


3969 


4021 


4072 


4124 4176 


4228 


52 


840 


924279 


924331 


924383 


924434 


924486 


924538 


9^4589 


924641 


924693 


924744! 52 


1 


4796 


4848 


4899 


4951 


6003 


6054 


6106 


6157 


5209 


6261 62 


2 


5312 


5364 


5415 


6467 


5518 


6570 


6621 


6673 


5725 


677C|52 


3 


5828 


6879 


6931 


6982 


6034 


6085 


6137 


6188 


6240 


6291 


51 


4 


6342 


6394 


6445 


6497 


6548 


6600 


6651 


6702 


6754 


6805 


51 


5 


6857 


6908 


6959 


7011 


7062 


7114 


7165 


7216 


7268 


7319 


61 


6 


7370 


7422 


7473 


7524 


757tt 


7627 


7678 


7730 


7781 


7832 


61 


7 


7883 


7935 


7986 


8037 


8088 


8140 


8191 


8242 


8293 


8345 


51 


8 8396 


8447 


8498- 


8549 


. 8601 


8652 


8703 


8754 


8805 


8857 


51 


9 


8908 


8959 


9010 


9061 


9112 


9163 


9215 


9266 


9317 


9368 61 


850 


929419 


929470 


929521 


929572 


929623 


929674 


929725|92 


929827 


929879 


51 


1 


9930 


9981 


930032 


930083 


930134 


930185 


930236 


930287 


930338 


930389 


51 


2 


930440 


930491 


0542 


0592 


0643 


0694 


0745 


0796 


0847 


0898 


51 


3 


0949 


1000 


1051 


1102 


1153 


1204 


1254 


1305 


1356 


1407 


51 


4 


1458 


1509 


1560 


1610 


1661 


1712 


1763 


1814 


1865 


1915 


51 


5 


1966 


2017 


2068 


2118 


2169 


2220 


2271 


2322 


2372 


2423 


51 


6 


2474 


2524 


2576 


2626 


2677 


2727 


2778 


2829 


2879 


2930 


51 


7 


2981 


3031 


3082 


3133 


3183 


3234 


3285 


3335 


3386 


3437 


51 


8 


3487 


3538 


3589 


3639 


3690 


3740 


3791 


3841 


3892 


3943 


51 


9 


3993 


4044 


4094 


4145 


4195 


4246 


4296 


4347 


4397 


4448 


51 


860 


934498 


934549 


934599 934050 


934700 


934751 


934801 


934852,93490 


934953 


50 


1 


6003 


6054 


6104 


5154 


6205 


6265 


5306 


5356 


5406 


5457 


60 


2 


5507 


5558 


6608 


5658 


5709 


5759 


6809 


6860 


5910 


5960 


50 


3 


6011 


6061 


6111 


6162 


6212 


6262 


6313 


6363 


6413 


6463 


50 


4 


6514 


6564 


6614 


6665 


6715 


6765 


6815 


6865 


6916 


6966 


50 


5 


7016 


7066 


7117 


7167 


7217 


7267 


7317 


7367 


7418 


7468 


50 


G 


7518 


7568 


7618 


7668 


7718 


7769 


7819 


7869 


7919 


7969 


50 


7 


8019 


8069 


8119 


8169 


8219 


8269 


8320 


8370 


8420 


8470 


50 


8 


8520 


8570 


8620 


8670 


8720 


8770 


8820 


8870 


8920 


8970 


60 


9 


9020 


9070 


9120 


9170 


9220 


9270 


9320 


9369 


9419 


9469 


50 


870 


939519 


969569 


939619 


939669939719 


939769 


939819 


93986919; 


939968 


60 


1 


940018 


940068 


940118 


940168 940218 


940267 


940317 


940367 


940417 


940467 


50 


2 


0616 


0566 


0616 


0666 


0716 


0765 


0815 


0865 


0915 


0964 


60 


3 


1014 


1064 1114 


1163 


1213 


1263 


1313 


1362 


1412 


1462 


50 


4 


1611 


1661 1611 


1660 


1710 


1760 


1809 


1859 


1909 


1958 


50 


5 


2008 


2058 2107 


2167 


2207 


2256 


2306 


2355 


2405 


2455 


50 


6 


'2504 


25541 2603 


2653 


2702 


2762 


2801 


2851 


2901 


2950 


50 


7 


3000 


3049 3099 


3148 


3198 


3247 


3297 


3346 


3396 


3445 


49 


8 


3445 


3544| 3593 


3643 


3692 


3742. 3791 


3841 


3890 


3939, 49 


9 


3989 40381 4088 


4137 


4186 


4236 1 4285 


4335 


4384 1 4433 49 


N. 





1 | 2 


3 


4 II 


6 


7 


8 9 ID. 



OF NUMBERS 



15 



gj o 


1 


2 


3 


4 


3 


6 


7 8 


9 


TT 


880 


944483 


944532 


944581 


944631 


944680 


944729 


944779 


944828 


944877 


944927 


49 


1 


4976 


5025 


5074 


6124 


5173 


5222 


6272 


5321 


6370 


64J 9 


49 


2 


5469 


5518 


5567 


5616 


5665 


5715 


5764 


5813 


6862 


69 v 12 


49 


3 


5961 


6010 


6059 


6108 


6157 


6207 


6256 


6305 


6354 


6403 


49 


4 


6452 


6501 


6551 


6600 


6649 


6698 


6747 


6796 


6845 


6894 


49 


6 


6943 


6992 


7041 


7090 


7140 


7189 


7238 


7287 


7336 


7385 


49 


6 


7434 


7483 


7532 


7561 


7630 


7679 


7728 


7777 


7826 


7875 


49 


7 


7924 


7973 


8022 


8070 


8119 


8168 


8217 


8266 


8315 


8364 


49 


8 


8413 


8462 


8511 


8560 


8609 


8657 


8706 


8755 


8804 


8853 


49 


9 


8902 


8951 


8999 


9048 


9097 


9146 


9195] 9244 


9292 


9341 


49 


890 


949390 


949439 


949488 


949536 


949585 949634 


949683 


949731 


949780 


949829 


49 


1 


9878 


9926 


9975 


950024 


950073 


950121 


950170 


950219 


950267 


950316 


49 


2 


950365 


950414 


950462 


0511 


0560 


0608 


0657 


0706 


0754 


0803 


49 


3 


0851 


0900 


0949 


0997 


1046 


1095 


1143 


1192 


1240 


1289 


49 


4 


1338 


1386 


1435 


1483 


1532 


1580 


1629 


1677 


1726 


1775 


49 


5 


1823 


1872 


1920 


1969 


2017 


2066 


2114 


2163 


2211 


2260 


48 


6 


2308 


2356 


2405 


2453 


2502 


2550 


2599 


2647 


2696 


2744 


48 


7 


2792 


2841 


2889 


2938 


2986 


3034 


3083 


3131 


3180 


3228 


48 


8 


3276 


3325 


3373 


3421 


3470 


3518 


3566 


3615 


3663 


3711 


48 


9 


3760 


3808 


3856 


3905 3953 


4001 


4049 


4098 4146 


4194 


48 


900 


954243 


9542? 1 


954339 95438 


954435 


954484 


954532 


954580 


954628 


954677 


48 


1 


4725 


4773 


4821 


4869 


4918 


4966 


6014 


5062 


5110 


5158 


48 


2 


5207 


5255 


6303 


5351 


5399 


5447 


6495 


5543 


6592 


5640 


48 


3 


5688 


5736 


5784 


5832 


6880 


6928 


5976 


6024 


6072 


6120 


48 


4 


6168 


6216 


6265 


6313 


6361 


6409 


6457 


6505 


6553 


6601 


48 


5 


6649 


6697 


6745 


6793 


6840 


6888 


6936 


6984 


7032 


7080 


48 


6 


7128 


7176 


7224 


7272 


7320 


7368 


7416 


7464 


7512 


7559 


48 


7 


7607 


7655 


7703 


7751 


7799 


7847 


7894 


7942 


7990 


8038 


48 


8 


8086 


8134 


8181 


8229 


8277 


8325 


8373 


8421 


8468 


8516 


48 


9 


8564 


8612 


8659 


8707 


8755 


8803 


8850 


8898 


8946 


8994 


48 


910 


959041 


959089 


959137 


959185 


959232 


959280 


959328 


959375 


959423 


959471 


48 


1 


9518 


9566 


9614 


9661 


9709 


9757 


9804 


9852 


9900 


9947 


48 


2 


9995 


960042 


960090 


960138 


960185 


960233 


960281 


960328 


960376 


960423 


48 


3 


960471 


0518 


0566 


0613 


066L 


0709 


0756 


0804 


0851 


0899 


48 


4 


0946 


0994 


1041 


1089 


1136 


1184 


1231 


1279 


1326 


1374 


48 


5 


1421 


1469 


1516 


1563 


1611 


1658 


1706 


1753 


1801 


1848 


47 


6 


1895 


1943 


1990 


2038 


2085 


2132 


2180 


2227 


2275 


2322 


47 


7 


2369 


2417 


2464 


2511 


2559 


2606 


2653 


2701 


2748 


2795 


47 


8 


2843 


2890 


2937 


2985 


3032 


3079 


3126 


3174 


3221 


3268 


47 


9 


3316 


3363 


3410 


3457 


3504 


3552 


3599 


3646 


3693 


3741 


47 


920 


963788 


963835 


963882 


963929 


963977 


964024 


964071 


964118 


964165 


964212 


47 


1 


4260 


4307 


4354 


4401 


4448 


4495 


4542 


4590 


4637 


4684 


47 


2 


4731 


4778 


4825 


4872 


4919 


4966 


6013 


5061 


6108 


5155 


47 


3 


5202 


6249 


5296 


5343 


5390 


5437 


5484 


5531 


6578 


6625 


47 


4 


5672 


6719 


5766 


6813 


6860 


5907 


5954 


6001 


6048 


6095 


47 


5 


6142 


6189 


6236 


6283 


6329 


6376 


6423 


6470 


6517 


6564 


47 


6 


6611 


6658 


6705 


6752 


6799 


6845 


6892 


6939 


C986 


7033 


47 


7 


7080 


7127 


7173 


7220 


7267 


7314 


7361 


7408 


7454 


7501 


47 


8 


7548 


7595 


7642 


7688 


7735 


7782 


7829 


7875 


7922 


7969 


47 


9 


8016 


8062 


8109 


8156 


8203 


6249 


8296 


8343 


8390 


8436 


47 


930 


968483 968531 


968576 


968623 


968670 


968716 


968763 96881 


968856 


968903 


47 


1 


8950 


8996 


9043 


9090 


9136 


9183 


9229 


9276 


9323 


9369 


47 


2 


9416 


9463 


9509 


9556 


9602 


9649 


9695 


9742 


9789 


9835 


47 


3 


9882 


9928 


9975 


970021 


970068 


970114 


970161 


970207 


970254 


970300 


47 


4 


970347 


970393 


970440 


0486 


0533 


0579 


0626 


0672 


0719 


0765 


46 


5 


0812 


0858 


0904 


0951 


0997 


1044 


1090 


1137 


1183 


1229 


46 


6 


1276 


1322 


1369 


1415 


1461 


1508 


1554 


1601 


1647 


1693 


46 


7 


1740 


1786 


1832 


1879 


1925 


1971 


2018 


2064 


2110 


2157 


46 


8 


2203 


2249 


2295 


2342 


2388 


2434 


2481 


2527 


2573 


2619 


46 


9 


2666 


2712 


2758 


2804 


2851 


2897 


2943 


2989 


3035 


3082 


46 


N. 





1 


4 


5 


« 


9 .d. 



16 



LOGARITHMS OF NUMBERS. 



N.| ) 2 3 


4 1 


6 6 


7 


E 


9 |I> 


94U-J73128 


973174 


973220 


9732'i6 


973313 


973359 9734U5 9' 


973497 


973543, 46 


1 


3590 


3636 


3682 


3728 


3774 


3820 


3866 


3913 


3959 


4005! 46 


2 


4051 


4097 


4143 


4189 


4235 


4281 


4327 


4374 


4420 


44661 46 


3 


4512 


4558 


4604 


4650 


4696 


4742 


4788 


4834 


4880 


4926 


46 


4 


4972 


6018 


5064 


5110 


5156 


5202 


6248 


6294 


6340 


6386 


46 


5 


5432 


6478 


5524 


5570 


5616 


5662 


6707 


6753 


6799 


6845 


46 


6 


6891 


6937 


5983 


6029 


6075 


6121 


6167 


6212 


6258 


6304 


46 


7 


6350 


6396 


6442 


6488 


6533 


6579 


6625 


6-371 


6717 


6763 


46 


8 


6808 


6854 


6900 


6946 


6992 


7037 


7083 


7129 


7175 


7220 


46 


9 


7266 


7312 


7358 


7403 


7449 


7495 7641 


7586 


7632 


7678 


46 


9'30 


977724 


9777.69 


977815 


977861 


977906 


977952 


977998 


978043 


978089 


978135 


*«< 


1 


,8,18] 
* 803 7 


8226 


8272 


8317* 


8*63 


8409 


mi 


8600 


8548 


S6S1 


4* 


2 


8683 


8728 


8774 


8819 


8865 
9321 


8911 


8956 


9002 


9047 


46 


3 


9093 


9138 


9184 


9230 


9275 


9366 


9412 


9457 


9503 


IV 


4 


9548 


9594 


9639 


9685 


9730 


9776 


9821 


9*67 


9912 


9958 


46 


5 


980003 


980049 


980094 


980140 


980185 


980231 


980276 


980322 


980367 


980412 


46 


6 


0458 


0503 


0549 


0594 


0640 


0685 


0730 


0776 


0821 


0867 


45 


7 


0912 


0957 


1003 


1048 


1093 


1139 


1184 


1229 


1275 


1320 


45 


8 


1366 


1411 


1456 


1501 


1547 


1592 


1637 


1683 


1728 


17T3 


45 


9 


1819 


1864 


1909 


1954 


2000 


2045 


2090 2135 


2181 


2226 


45 


960 


982271 


982316 


982362 


982407 


982452 


982497 


982543 


982588 


982633 


982678 


45 


1 


2723 


2769 


2814 


2859 


2904 


2949 


2994 


3040 


3085 


3130 


45 


2 


3175 


3220 


3265 


3310 


3356 


3401 


3446 


3491 


3536 


3581 


45 


3 


3626 


3671 


3716 


3762 


3807 


3852 


3897 


3942 


3987 


4032 


45 


4 


4077 


4122 


4167 


4212 


4257 


4302 


4347 


4392 


4437 


4482 


45 


5 


4527 


4572 


4617 


4662 


4707 


4762 


4797 


4842 


4887 


4932 


45 


6 


4977 


6022 


6067 


6112 


6167 


5202 


6247 


5292 


6337 


5382 


45 


7 


5426 


6471 


6516 


6561 


5606 


6651 


6696 


5741 


6786 


6830 


45 


8 


6875 


6920 


6965 


6010 


6055 


6100 


6144 


6189 


6234 


6279 


45 


9 


6324 


6369 


6413 


6458 


6503 


6548 


6593 


6637 


6682 


6727 


45 


970 


986772 91 


986861 


986906 


986951 


986996 


987040 


987085 


987130 


987176 


45 


1 


7219 


7264 


7309 


7353 


7398 


7443 


7488 


7532 


7577 


7622 


45 


2 


7666 


7711 


7756 


7800 


7845 


7890 


7934 


7979 


8024 


8068 


45 


3 


8113 


8157 


8202 


8247 


8291 


8336 


8381 


8425 


8470 


8514 


45 


4 


8559 


8604 


8648 


8693 


8737 


8782 


8826 


8871 


8916 


8960 


45 


6 


9005 


9049 


9094 


9138 


9183 


9227 


9272 


9316 


9361 


9405 


45 


6 


9450 


9494 


9539 


9583 


9628 


9672 


9717 


9761 


9806 


9850 


44 


7 


9895 


9939 


9983 


990028990072 


990117 


990161 


990206 


990250 


990294 


44 


8 


990339 


990383 


990428 


0472! 0516 


0561 


0605 


0650 


0694 


0738 


44 


9 


0783 


0827 


0871 


0916! 0960 


1004 


1049 


1093 


1137 


1182 


44 


^80 


991226 


991270,9! 


991359 


991403 


991448 


991492 


991536 


991580 


991625 


44 


1 


1669 


1713 


1758 


1802 


1846 


1890 


1935 


1979 


2023 


2067 


41 


'2 


2111 


2156 


2200 


2244 


2288 


2333 


2377 


2421 


2465 


2509 


44 


3 


2554 


2598 


2642 


2686 


2730 


2774 


2819 


2863 


2907 


2951 


44 


4 


2995 


3039 


3083 


3127 


3172 


3216 


3260 


3304 


3348 


3392 


44 


5 


3436 


3480 


3524 


3568 


3613 


3657 


3701 


3745 


3789 


3833 44 


6 


3877 


3921 


3965 


4009 


4053 


4097 


4141 


4185 


4229 


4273 


44 


.7 


4317 


4361 


4405 


4449 


4493 


4537 


4581 


4625 


4669 


4713 


44 


8 


4757 


4801 4845 


4889 


4933 


4977 


6021 


6065 


5108 


6152 


44 


9 

990 


5196 


6240| 6284 


6328 


5372 


6416 


6460 


6504 


6547 


6591 44 


995635 


9956791996723 99576 


995854 


995898 


995942 995986 


996030, 44 


1 


6074 


6117 


6161 


6205 


6249 


6293 6337 


6380 


6424 


6468i 44 


2 


6512 


6555 


6599 


6643 


6687 


6731 


6774 


6818 


6862 


6906 


44 


3 


6949 


6993 


7037 


7080 


7124 


7168 


7212 


7255 


7299 


7343 


44 


4 


7386 


7430 


7474 


7517 


7661 


7605 


7648 


7692 


7736 


7779 


44 


5 


7823 


7867 


7910 


7954 


7998 


8041 


8085 


8129 


8172 


8216 


44 


6 


8259 


8303 


8347 


8390 


8434 


8477 


8521 


8564 8608 


8652 


44 


7 


8695 


8739 


8782 


8826 


8869 


8913 


8956 


9000 


9043 


9087 


44 


8 


9131 


9174 


9218 


9261 


9305 


9348 


9392 


9435 


9479 


9522 


44 


9 9565 


9609 


9652 


9696 


9739 


9783 


9826 


9870 


9913 


9957 43 


nTT 


1 o 


1 


2 


4 


J & 


6 


7 


8 


9 ID. 



14 DAY USE 

RETURN TO DESK FROM WHT t BORROWED 



LOAN D 



This book is due on the la- 
or on the date to which r 
B „ no . Tel. No. 

Kenewals may be mad 
Kenewed books are 



Due.er 
— strK 




/- 
797965 <¥ 



►. 



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